7-1 Parabolas
For each equation, identify the vertex, focus,
axis of symmetry, and directrix. Then graph
the parabola.
2
1. (x – 3) = 12(y – 7)
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
2
vertically. The equation is in the form (x − h) = 4p
(y − k), so h = 3 and k = 7. Because 4p = 12 and p
= 3, the graph opens up.Use the values of h, k, and
p to determine the characteristics of the parabola.
vertex: (h, k) = (3, 7)
focus: (h, k + p ) = (3, 10)
directrix: y = k – p or 4
axis of symmetry: x = h or 3
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
2. (x + 1)2 = −12(y – 6)
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
2
vertically. The equation is in the form (x − h) =
4p (y − k), so h = –1 and k = 6. Because 4p = –12
and p = –3, the graph opens down.Use the values
of h, k, and p to determine the characteristics of
the parabola.
vertex: (h, k) = (–1, 6)
focus: (h, k + p ) = (–1, 3)
directrix: y = k – p or 9
axis of symmetry: x = h or –1
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
3. (y – 4)2 = 20(x + 2)
2. (x + 1)2 = −12(y – 6)
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
2
vertically. The equation is in the form (x − h) =
4p (y − k), so h = –1 and k = 6. Because 4p = –12
and p = –3, the graph opens down.Use the values
of h, k, and p to determine the characteristics of
the parabola.
vertex: (h, k) = (–1, 6)
focus: (h, k + p ) = (–1, 3)
directrix: y = k – p or 9
axis of symmetry: x = h or –1
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
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SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
2
horizontally. The equation is in the form (y − k) =
4p (x − h), so h = –2 and k = 4. Because 4p = 20
and p = 5, the graph opens to the right. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–2, 4)
focus: (h + p , k) = (3, 4)
directrix: x = h – p or –7
axis of symmetry: y = k or 4
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
Page 1
7-1 Parabolas
3. (y – 4)2 = 20(x + 2)
4. −1(x + 7) = (y + 5)2
SOLUTION:
SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
2
horizontally. The equation is in the form (y − k) =
4p (x − h), so h = –2 and k = 4. Because 4p = 20
and p = 5, the graph opens to the right. Use the
values of h, k, and p to determine the
characteristics of the parabola.
The equation is in standard form and the squared
term is y, which means that the parabola opens
horizontally. The equation is in the form 4p (x − h) =
2
(y − k) , so h = –7 and k = –5. Because 4p = –1
and p = −0.25, the graph opens to the left. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–2, 4)
focus: (h + p , k) = (3, 4)
directrix: x = h – p or –7
axis of symmetry: y = k or 4
vertex: (h, k) = (–7, –5)
focus: (h + p , k) = (–7.25, –5)
directrix: x = h – p or –6.75
axis of symmetry: y = k or –5
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
4. −1(x + 7) = (y + 5)2
SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
horizontally. The equation is in the form 4p (x − h) =
2
(y − k) , so h = –7 and k = –5. Because 4p = –1
and p = −0.25, the graph opens to the left. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–7, –5)
focus: (h + p , k) = (–7.25, –5)
directrix: x = h – p or –6.75
axis of symmetry: y = k or –5
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
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5. (x + 8)2 = 8(y – 3)
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
2
vertically. The equation is in the form (x − h) = 4p
(y − k), so h = –8 and k = 3. Because 4p = 8 and p
= 2, the graph opens up. Use the values of h, k, and
p to determine the characteristics of the parabola.
vertex: (h, k) = (–8, 3)
focus: (h, k + p ) = (–8, 5)
directrix: y = k – p or 1
axis of symmetry: x = h or –8
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
Page 2
7-1 Parabolas
5. (x + 8)2 = 8(y – 3)
6. −40(x + 4) = (y – 9)2
SOLUTION:
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
2
vertically. The equation is in the form (x − h) = 4p
(y − k), so h = –8 and k = 3. Because 4p = 8 and p
= 2, the graph opens up. Use the values of h, k, and
p to determine the characteristics of the parabola.
The equation is in standard form and the squared
term is y, which means that the parabola opens
horizontally. The equation is in the form 4p (x − h) =
2
(y − k) , so h = –4 and k = 9. Because 4p = –40
and p = –10, the graph opens to the left. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–8, 3)
focus: (h, k + p ) = (–8, 5)
directrix: y = k – p or 1
axis of symmetry: x = h or –8
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
6. −40(x + 4) = (y – 9)2
SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
horizontally. The equation is in the form 4p (x − h) =
2
(y − k) , so h = –4 and k = 9. Because 4p = –40
and p = –10, the graph opens to the left. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–4, 9)
focus: (h + p , k) = (–14, 9)
directrix: x = h – p or 6
axis of symmetry: y = k or 9
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
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vertex: (h, k) = (–4, 9)
focus: (h + p , k) = (–14, 9)
directrix: x = h – p or 6
axis of symmetry: y = k or 9
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
7. (y + 5)2 = 24(x − 1)
SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
2
horizontally. The equation is in the form (y − k) =
4p (x − h), so h = 1 and k = –5. Because 4p = 24
and p = 6, the graph opens to the right. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (1, –5)
focus: (h + p , k) = (7, –5)
directrix: x = h – p or –5
axis of symmetry: y = k or –5
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
Page 3
7-1 Parabolas
7. (y + 5)2 = 24(x − 1)
8. 2(y + 12) = (x – 6)2
SOLUTION:
SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
2
horizontally. The equation is in the form (y − k) =
4p (x − h), so h = 1 and k = –5. Because 4p = 24
and p = 6, the graph opens to the right. Use the
values of h, k, and p to determine the
characteristics of the parabola.
The equation is in standard form and the squared
term is x, which means that the parabola opens
vertically. The equation is in the form 4p (y − k) =
2
(x − h) , so h = 6 and k = –12. Because 4p = 2 and
vertex: (h, k) = (1, –5)
focus: (h + p , k) = (7, –5)
directrix: x = h – p or –5
axis of symmetry: y = k or –5
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
p=
, the graph opens up. Use the values of h, k,
and p to determine the characteristics of the
parabola.
vertex: (h, k) = (6, –12)
focus: (h + p , k) = (6, –11.5)
directrix: y = k – p or –12.5
axis of symmetry: x = h or 6
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
8. 2(y + 12) = (x – 6)2
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
vertically. The equation is in the form 4p (y − k) =
2
(x − h) , so h = 6 and k = –12. Because 4p = 2 and
p=
, the graph opens up. Use the values of h, k,
and p to determine the characteristics of the
parabola.
vertex: (h, k) = (6, –12)
focus: (h + p , k) = (6, –11.5)
directrix: y = k – p or –12.5
axis of symmetry: x = h or 6
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
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9. −4(y +2) = (x + 8)2
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
vertically. The equation is in the form 4p (y − k) =
2
(x − h) , so h = –8 and k = –2. Because 4p = –4
and p = –1, the graph opens down. Use the values
of h, k, and p to determine the characteristics of
the parabola.
vertex: (h, k) = (–8, –2)
focus: (h + p , k) = (–8, –3)
directrix: y = k – p or –1
axis of symmetry: x = h or –8
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
Page 4
7-1 Parabolas
9. −4(y +2) = (x + 8)2
10. 10(x + 11) = (y + 3)2
SOLUTION:
SOLUTION:
The equation is in standard form and the squared
term is x, which means that the parabola opens
vertically. The equation is in the form 4p (y − k) =
2
(x − h) , so h = –8 and k = –2. Because 4p = –4
and p = –1, the graph opens down. Use the values
of h, k, and p to determine the characteristics of
the parabola.
The equation is in standard form and the squared
term is y, which means that the parabola opens
horizontally. The equation is in the form 4p (x − h) =
2
(y − k) , so h = –11 and k = –3. Because 4p = 10
and p = 2.5, the graph opens to the right. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–8, –2)
focus: (h + p , k) = (–8, –3)
directrix: y = k – p or –1
axis of symmetry: x = h or –8
vertex: (h, k) = (–11, –3)
focus: (h + p , k) = (–8.5, –3)
directrix: x = h – p or –13.5
axis of symmetry: y = k or –3
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
10. 10(x + 11) = (y + 3)2
SOLUTION:
The equation is in standard form and the squared
term is y, which means that the parabola opens
horizontally. The equation is in the form 4p (x − h) =
2
(y − k) , so h = –11 and k = –3. Because 4p = 10
and p = 2.5, the graph opens to the right. Use the
values of h, k, and p to determine the
characteristics of the parabola.
vertex: (h, k) = (–11, –3)
focus: (h + p , k) = (–8.5, –3)
directrix: x = h – p or –13.5
axis of symmetry: y = k or –3
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
general shape of the curve.
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11. SKATEBOARDING A group of high school
students designing a half–pipe have decided that
the ramps, or transitions, could be obtained by
splitting a parabola in half. A parabolic cross
2
section of the ramps can be modeled by x = 8(y −
2), where the values of x and y are measured in
feet. Where is the focus of the parabola in relation
to the ground if the ground represents the directrix?
SOLUTION:
Because the x–term is squared and p is positive,
the parabola opens up and the focus is located at
(h, k + p ). The equation is provided in standard
form, and h = 0 and k = 2. Because 4p = 8, p is 2.
So, the location of the focus is (0, 2 + 2) or (0, 4).
Therefore, the focus is 4 feet above the ground.
12. COMMUNICATION The cross section of a
satellite television dish has a parabolic shape that
focuses the satellite signals onto a receiver located
at the focus of the parabola. The parabolic cross
2
Page 5
section can be modeled by (x – 6) = 12(y – 10),
where the values of x and y are measured in
inches. Where is the receiver located in relation to
7-1
the parabola opens up and the focus is located at
(h, k + p ). The equation is provided in standard
form, and h = 0 and k = 2. Because 4p = 8, p is 2.
So, the location of the focus is (0, 2 + 2) or (0, 4).
Parabolas
Therefore, the focus is 4 feet above the ground.
12. COMMUNICATION The cross section of a
satellite television dish has a parabolic shape that
focuses the satellite signals onto a receiver located
at the focus of the parabola. The parabolic cross
2
section can be modeled by (x – 6) = 12(y – 10),
where the values of x and y are measured in
inches. Where is the receiver located in relation to
this particular cross section?
SOLUTION:
The receiver is located at the focus of the satellite
dish. Because the x–term is squared and p is
positive, the parabola opens up and the focus is
located at (h, k + p ). The equation is provided in
standard form, and h = 6 and y = 10. Because 4p =
12, p is 3. So, the location of the focus is (6, 10 + 3)
or (6, 13). The location of the focus for the satellite
dish is (6, 13). Therefore, the receiver is 3 inches
above the vertex of the satellite dish.
13. BOATING As a speed boat glides through the
water, it creates a wake in the shape of a
parabola. The vertex of this parabola meets with
the stern of the boat. A swimmer on a wakeboard,
attached by a piece of rope, is being pulled by the
boat. When he is directly behind the boat, he is
positioned at the focus of the parabola. The
parabola formed by the wake can be modeled using
standard form, and h = 6 and y = 10. Because 4p =
12, p is 3. So, the location of the focus is (6, 10 + 3)
or (6, 13). The location of the focus for the satellite
dish is (6, 13). Therefore, the receiver is 3 inches
above the vertex of the satellite dish.
13. BOATING As a speed boat glides through the
water, it creates a wake in the shape of a
parabola. The vertex of this parabola meets with
the stern of the boat. A swimmer on a wakeboard,
attached by a piece of rope, is being pulled by the
boat. When he is directly behind the boat, he is
positioned at the focus of the parabola. The
parabola formed by the wake can be modeled using
2
y − 180x + 10y + 565 = 0, where x and y are
measured in feet.
a. Write the equation in standard form.
b. How long is the length of rope attaching the
swimmer to the stern of the boat?
SOLUTION:
a.
2
y − 180x + 10y + 565 = 0, where x and y are
measured in feet.
a. Write the equation in standard form.
b. How long is the length of rope attaching the
swimmer to the stern of the boat?
SOLUTION:
a.
b.The equation in standard form has y as the
squared term, which means that the parabola opens
horizontally. Because 4p = 180, p = 45 and the
graph opens to the right. The equation is in the form
2
(y − k) = 4p (x − h), so h = 3 and k = –5. Since
the stern is located at the vertex of the parabola
formed, it is at the point (h, k) or (3, –5).
The swimmer is at the focus, located at (h + p , k),
which is (3 + 45, –5) or (48, –5). The distance the
swimmer is from the stern of the boat represents
the length of rope needed to attach the swimmer to
the stern.
Using the distance formula, the distance between
these two points is
or 45.
Therefore, the length of rope attaching the
swimmer to the stern of the boat is 45 feet.
b.The equation in standard form has y as the
squared term, which means that the parabola opens
horizontally.
Because
4p = 180, p = 45 and the
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graph opens to the right. The equation is in the form
2
(y − k) = 4p (x − h), so h = 3 and k = –5. Since
14. BASEBALL During Philadelphia Phillies baseball
games, the team’s mascot, The Phanatic, launches
Page 6
hot dogs into the stands. The launching device
propels the hot dogs into the air at an initial velocity
of 64 feet per second. A hot dog’s distance y
2
these two points is
7-1
or 45.
Therefore,
the
length
of
rope
attaching
the
Parabolas
swimmer to the stern of the boat is 45 feet.
14. BASEBALL During Philadelphia Phillies baseball
games, the team’s mascot, The Phanatic, launches
hot dogs into the stands. The launching device
propels the hot dogs into the air at an initial velocity
of 64 feet per second. A hot dog’s distance y
above ground after x seconds can be illustrated by
2
− k) = (x − h) , so h = 2 and k = 70. The
maximum height a hot dog can reach is located at
the vertex of the parabola formed. It is at the point
(h, k) or (2, 70). Therefore, the maximum height a
hot dog can reach is 70 feet.
Write each equation in standard form. Identify
the vertex, focus, axis of symmetry, and
directrix. Then graph the parabola.
15. x2 − 17 = 8y + 39
SOLUTION:
2
y = −16x + 64x + 6.
a. Write the equation in standard form.
b. What is the maximum height of a propelled hot
dog?
Because the x–term is squared and p = 2, the
graph opens up. Use the standard form to
determine the characteristics of the parabola.
SOLUTION:
a.
vertex: (h, k) = (0, –7)
focus: (h, k + p ) = (0, –5)
directrix: y = k – p or –9
axis of symmetry: x = h or 0
b.The equation in standard form has x as the
squared term, which means that the parabola opens
vertically. Because
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
and the
graph opens down. The equation is in the form 4p (y
2
2
− k) = (x − h) , so h = 2 and k = 70. The
maximum height a hot dog can reach is located at
the vertex of the parabola formed. It is at the point
(h, k) or (2, 70). Therefore, the maximum height a
hot dog can reach is 70 feet.
Write each equation in standard form. Identify
the vertex, focus, axis of symmetry, and
directrix. Then graph the parabola.
15. x2 − 17 = 8y + 39
16. y 2 + 33 = −8x − 23
SOLUTION:
SOLUTION:
Because the x–term is squared and p = 2, the
graph opens up. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (0, –7)
focus: (h, k + p ) = (0, –5)
directrix:
= k – pbyor
–9
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axis of symmetry: x = h or 0
Graph the vertex, focus, axis, and directrix of the
Because the y–term is squared and p = –2, the
graph opens to the left. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (–7, 0)
focus: (h + p , k) = (–9, 0)
directrix: x = h – p or –5
axis of symmetry: y = k or 0
Page 7
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
7-1 Parabolas
16. y 2 + 33 = −8x − 23
SOLUTION:
17. 3x2 + 72 = −72y
SOLUTION:
Because the y–term is squared and p = –2, the
graph opens to the left. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (–7, 0)
focus: (h + p , k) = (–9, 0)
directrix: x = h – p or –5
axis of symmetry: y = k or 0
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
Because the x–term is squared and p = −6, the
graph opens down. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (0, –1)
focus: (h, k + p ) = (0, –7)
directrix: y = k – p or 5
axis of symmetry: x = h or 0
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
17. 3x2 + 72 = −72y
SOLUTION:
18. −12y + 10 = x2 – 4x + 14
SOLUTION:
Because the x–term is squared and p = −6, the
graph opens down. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (0, –1)
focus: (h, k + p ) = (0, –7)
directrix: y = k – p or 5
axis of symmetry: x = h or 0
Graph
the- Powered
vertex, by
focus,
axis,
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and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
Because the x–term is squared and p = −3, the
graph opens down. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (2, 0)
focus: (h, k + p ) = (2, −3)
directrix: y = k – p or 3
axis of symmetry: x = h or 2
Page 8
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
7-1 Parabolas
18. −12y + 10 = x2 – 4x + 14
SOLUTION:
Because the x–term is squared and p = −3, the
graph opens down. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (2, 0)
focus: (h, k + p ) = (2, −3)
directrix: y = k – p or 3
axis of symmetry: x = h or 2
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
19. 60x – 80 = 3y 2 + 100
SOLUTION:
Because the y–term is squared and p = 5, the
graph opens to the right. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (3, 0)
focus: (h + p , k) = (8, 0)
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directrix:
= h – pbyorCognero
−2
axis of symmetry: y = k or 0
19. 60x – 80 = 3y 2 + 100
SOLUTION:
Because the y–term is squared and p = 5, the
graph opens to the right. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (3, 0)
focus: (h + p , k) = (8, 0)
directrix: x = h – p or −2
axis of symmetry: y = k or 0
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
20. −33 = x2 – 12y – 6x
SOLUTION:
Because the x–term is squared and p = 3, the
graph opens up. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (3, 2)
focus: (h, k + p ) = (3, 5)
directrix: y = k – p or −1
Page 9
7-1 Parabolas
20. −33 = x2 – 12y – 6x
SOLUTION:
21. −72 = 2y 2 – 16y − 20x
SOLUTION:
Because the x–term is squared and p = 3, the
graph opens up. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (3, 2)
focus: (h, k + p ) = (3, 5)
directrix: y = k – p or −1
axis of symmetry: x = h or 3
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
Because the y–term is squared and p = 2.5, the
graph opens to the right. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (2, 4)
focus: (h + p , k) = (4.5, 4)
directrix: x = h – p or −0.5
axis of symmetry: y = k or 4
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
21. −72 = 2y 2 – 16y − 20x
SOLUTION:
22. y 2 + 21 = −20x – 6y − 68
SOLUTION:
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Page 10
7-1 Parabolas
22. y 2 + 21 = −20x – 6y − 68
23. x2 – 18y + 12x = 126
SOLUTION:
SOLUTION:
Because the y–term is squared and p = −5, the
graph opens to the left. Use the standard form to
determine the characteristics of the parabola.
Because the x–term is squared and p = 4.5, the
graph opens up. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (−4, −3)
focus: (h + p , k) = (−9, −3)
directrix: x = h – p or 1
axis of symmetry: y = k or −3
vertex: (h, k) = (−6, −9)
focus: (h, k + p ) = (−6, −4.5)
directrix: y = k – p or −13.5
axis of symmetry: x = h or –6
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
2
23. x – 18y + 12x = 126
24. −34 = 2x2 + 20x + 8y
SOLUTION:
SOLUTION:
Because the x–term is squared and p = 4.5, the
graph opens up. Use the standard form to
determine the characteristics of the parabola.
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vertex: (h, k) = (−6, −9)
focus: (h, k + p ) = (−6, −4.5)
directrix: y = k – p or −13.5
Page 11
7-1 Parabolas
24. −34 = 2x2 + 20x + 8y
SOLUTION:
25. LIGHTING Stadium lights at an athletic field need
to reflect light at maximum intensity. The bulb
should be placed at the focal point of the parabolic
globe surrounding it. If the shape of the globe is
2
given by x = 36y, where x and y are in inches,
how far from the vertex of the globe should the
bulb be placed for maximum light?
SOLUTION:
Because the x–term is squared and p = −1, the
graph opens down. Use the standard form to
determine the characteristics of the parabola.
vertex: (h, k) = (−5, 2)
focus: (h, k + p ) = (−5, 1)
directrix: y = k – p or 3
axis of symmetry: x = h or −5
Graph the vertex, focus, axis, and directrix of the
parabola. Then make a table of values to graph the
curve. The curve should be symmetric about the
axis of symmetry.
The bulb is placed at the focus of the parabola.
Because the x–term is squared and p is positive,
the parabola opens up and the focus is located at
(h, k + p ). The equation is provided in standard
form, and h = 0 and k = 0. Because 4p = 36, p is 9.
So, the location of the focus is (0, 0 + 9) or (0, 9).
The vertex of the globe is located at (0, 0). The
distance from the focus to the vertex is 9 − 0 or 9.
Therefore, the bulb should be placed 9 inches from
the vertex of the globe.
Write an equation for and graph a parabola
with the given focus F and vertex V.
26. F(−9, −7), V(−9, −4)
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is −7 − (−4) or −3. Because
p is negative, the graph opens down.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−3)[y – (−4)] = [x – (−9)]2
−12(y + 4) = (x + 9)2
2
The standard form of the equation is (x + 9) = −12
(y + 4).
Graph the vertex and focus. Then make a table of
values to graph the parabola.
25. LIGHTING Stadium lights at an athletic field need
to reflect light at maximum intensity. The bulb
should be placed at the focal point of the parabolic
globe surrounding it. If the shape of the globe is
2
given by x = 36y, where x and y are in inches,
how far from the vertex of the globe should the
bulb be placed for maximum light?
SOLUTION:
eSolutions
Manual
The
bulb-isPowered
placedbyatCognero
the focus
of the parabola.
Because the x–term is squared and p is positive,
the parabola opens up and the focus is located at
Page 12
27. F(2, −1), V(−4, −1)
SOLUTION:
7-1
So, the location of the focus is (0, 0 + 9) or (0, 9).
The vertex of the globe is located at (0, 0). The
distance from the focus to the vertex is 9 − 0 or 9.
Therefore, the bulb should be placed 9 inches from
Parabolas
the vertex of the globe.
Write an equation for and graph a parabola
with the given focus F and vertex V.
26. F(−9, −7), V(−9, −4)
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is −7 − (−4) or −3. Because
p is negative, the graph opens down.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−3)[y – (−4)] = [x – (−9)]2
−12(y + 4) = (x + 9)2
2
The standard form of the equation is (x + 9) = −12
(y + 4).
27. F(2, −1), V(−4, −1)
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is 2 – (−4) or 6. Because
p is positive, the graph opens to the right.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(6)[x – (−4)] = [y – (−1)]2
24(x + 4) = (y + 1)2
2
The standard form of the equation is (y + 1) = 24
(x + 4).Graph the vertex and focus. Then make a
table of values to graph the parabola.
Graph the vertex and focus. Then make a table of
values to graph the parabola.
28. F(−3, −2), V(1, −2)
SOLUTION:
27. F(2, −1), V(−4, −1)
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is 2 – (−4) or 6. Because
p is positive, the graph opens to the right.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(6)[x – (−4)] = [y – (−1)]2
24(x + 4) = (y + 1)2
2
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is −3 − 1 or −4. Because
p is negative, the graph opens to the left.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−4)(x – 1) = [y – (−2)]2
−16(x − 1) = (y + 2)2
2
The standard form of the equation is (y + 2) = −16
(x − 1).Graph the vertex and focus. Then make a
table of values to graph the parabola.
The standard form of the equation is (y + 1) = 24
(x + 4).Graph the vertex and focus. Then make a
table of values to graph the parabola.
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Page 13
7-1 Parabolas
28. F(−3, −2), V(1, −2)
29. F(−3, 4), V(−3, 2)
SOLUTION:
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is −3 − 1 or −4. Because
p is negative, the graph opens to the left.
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is 4 − 2 or 2. Because p is
positive, the graph opens up.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−4)(x – 1) = [y – (−2)]2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(2)(y – 2) = [x − (−3)]2
−16(x − 1) = (y + 2)2
8(y − 2) = (x + 3)2
2
The standard form of the equation is (y + 2) = −16
(x − 1).Graph the vertex and focus. Then make a
table of values to graph the parabola.
2
The standard form of the equation is (x + 3) = 8(y
– 2). Graph the vertex and focus. Then make a
table of values to graph the parabola.
30. F(−2, −4), V(−2, −5)
29. F(−3, 4), V(−3, 2)
SOLUTION:
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is 4 − 2 or 2. Because p is
positive, the graph opens up.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(2)(y – 2) = [x − (−3)]2
8(y − 2) = (x + 3)2
2
The standard form of the equation is (x + 3) = 8(y
– 2). Graph the vertex and focus. Then make a
table of values to graph the parabola.
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Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is −4 − (−5) or 1. Because p
is positive, the graph opens up.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(1)[y − (−5)] = [x − (−2)]2
(y + 5) = (x + 2)2
2
The standard form of the equation is (x + 2) = 4(y
+ 5). Graph the vertex and focus. Then make a
table of values to graph the parabola.
Page 14
7-1 Parabolas
30. F(−2, −4), V(−2, −5)
31. F(−1, 4), V(7, 4)
SOLUTION:
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is −4 − (−5) or 1. Because p
is positive, the graph opens up.
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is −1 − 7 or −8. Because
p is negative, the graph opens to the left.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(1)[y − (−5)] = [x − (−2)]2
(y + 5) = (x + 2)2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−8)(x − 7) = (y – 4)2
−32(x − 7) = (y − 4)2
2
The standard form of the equation is (x + 2) = 4(y
+ 5). Graph the vertex and focus. Then make a
table of values to graph the parabola.
2
The standard form of the equation is (y – 4) = −32
(x – 7). Graph the vertex and focus. Then make a
table of values to graph the parabola.
31. F(−1, 4), V(7, 4)
32. F(14, −8), V(7, −8)
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is −1 − 7 or −8. Because
p is negative, the graph opens to the left.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−8)(x − 7) = (y – 4)2
−32(x − 7) = (y − 4)2
2
The standard form of the equation is (y – 4) = −32
(x – 7). Graph the vertex and focus. Then make a
table of values to graph the parabola.
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SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is 14 – 7 or 7. Because p
is positive, the graph opens to the right.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(7)(x – 7) = [y – (−8)]2
28(x − 7) = (y + 8)2
2
The standard form of the equation is (y + 8) = 28
(x – 7). Graph the vertex and focus. Then make a
table of values to graph the parabola.
Page 15
7-1 Parabolas
32. F(14, −8), V(7, −8)
33. F(1, 3), V(1, 6)
SOLUTION:
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is 14 – 7 or 7. Because p
is positive, the graph opens to the right.
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is 3 − 6 or −3. Because p is
negative, the graph opens down.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(7)(x – 7) = [y – (−8)]2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−3)(y – 6) = (x – 1)2
28(x − 7) = (y + 8)2
−12(y − 6) = (x − 1)2
2
2
The standard form of the equation is (y + 8) = 28
(x – 7). Graph the vertex and focus. Then make a
table of values to graph the parabola.
The standard form of the equation is (x – 1) = −12
(y – 6). Graph the vertex and focus. Then make a
table of values to graph the parabola.
34. F(−4, 9), V(−2, 9)
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is −4 – (−2) or −2.
Because p is negative, the graph opens to the left.
33. F(1, 3), V(1, 6)
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is 3 − 6 or −3. Because p is
negative, the graph opens down.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−3)(y – 6) = (x – 1)2
−12(y − 6) = (x − 1)2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−2)[x – (−2)] = (y – 9)2
−8(x + 2) = (y − 9)2
2
The standard form of the equation is (y – 9) = −8
(x + 2). Graph the vertex and focus. Then make a
table of values to graph the parabola.
2
The standard form of the equation is (x – 1) = −12
(y – 6). Graph the vertex and focus. Then make a
table of values to graph the parabola.
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Page 16
35. F(8, −3), V(8, −7)
7-1 Parabolas
34. F(−4, 9), V(−2, 9)
35. F(8, −3), V(8, −7)
SOLUTION:
SOLUTION:
Because the focus and vertex share the same y–
coordinate, the graph is horizontal. The focus is (h
+ p , k), so the value of p is −4 – (−2) or −2.
Because p is negative, the graph opens to the left.
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is −3 − (−7) or 4. Because p
is positive, the graph opens up.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−2)[x – (−2)] = (y – 9)2
−8(x + 2) = (y − 9)2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(4)[y – (−7)] = (x – 8)2
16(y + 7) = (x − 8)2
2
The standard form of the equation is (y – 9) = −8
(x + 2). Graph the vertex and focus. Then make a
table of values to graph the parabola.
35. F(8, −3), V(8, −7)
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is −3 − (−7) or 4. Because p
is positive, the graph opens up.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(4)[y – (−7)] = (x – 8)2
16(y + 7) = (x − 8)2
2
The standard form of the equation is (x – 8) = 16
(y + 7). Graph the vertex and focus. Then make a
table of values to graph the parabola.
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2
The standard form of the equation is (x – 8) = 16
(y + 7). Graph the vertex and focus. Then make a
table of values to graph the parabola.
Write an equation for and graph each parabola
with focus F and the given characteristics.
36. F(3, 3); opens up; contains (23, 18)
SOLUTION:
Because the parabola opens up, the vertex is (3, 3
− p ). Use the standard form of the equation of a
vertical parabola and the point (23, 18) to find the
equation.
4p (y – k) = (x – h)2
4p [18 – (3 − p )] = (23 – 3)2
4p (15 + p ) = 400
2
4p + 60p = 400
2
p + 15p − 100 = 0
(p + 20)(p − 5) = 0
p = −20 or 5
Because the parabola opens up, the value of p must
be positive. Therefore, p = 5. The vertex is (3, −2),
2
and the standard form of the equation is (x – 3) =
20(y + 2). Use a table of values to graph the
parabola.
Page 17
7-1 Parabolas
Write an equation for and graph each parabola
with focus F and the given characteristics.
36. F(3, 3); opens up; contains (23, 18)
SOLUTION:
Because the parabola opens up, the vertex is (3, 3
− p ). Use the standard form of the equation of a
vertical parabola and the point (23, 18) to find the
equation.
4p (y – k) = (x – h)2
4p [18 – (3 − p )] = (23 – 3)2
4p (15 + p ) = 400
2
4p + 60p = 400
2
p + 15p − 100 = 0
(p + 20)(p − 5) = 0
p = −20 or 5
Because the parabola opens up, the value of p must
be positive. Therefore, p = 5. The vertex is (3, −2),
2
and the standard form of the equation is (x – 3) =
20(y + 2). Use a table of values to graph the
parabola.
37. F(1, 2); opens down; contains (7, 2)
SOLUTION:
Because the parabola opens down, the vertex is (1,
2 − p ). Use the standard form of the equation of a
vertical parabola and the point (7, 2) to find the
equation.
4p (y – k) = (x – h)2
4p [2 – (2 − p )] = (7 – 1)2
4p (p ) = 36
2
p =9
p = ±3
Because the parabola opens down, the value of p
must be negative. Therefore, p = −3. The vertex is
(1, 5), and the standard form of the equation is (x –
2
1) = −12(y − 5). Use a table of values to graph the
parabola.
38. F(11, 4); opens right; contains (20, 16)
SOLUTION:
37. F(1, 2); opens down; contains (7, 2)
SOLUTION:
Because the parabola opens down, the vertex is (1,
2 − p ). Use the standard form of the equation of a
vertical parabola and the point (7, 2) to find the
equation.
4p (y – k) = (x – h)2
4p [2 – (2 − p )] = (7 – 1)2
4p (p ) = 36
2
p =9
p = ±3
Because the parabola opens down, the value of p
must be negative. Therefore, p = −3. The vertex is
(1, 5), and the standard form of the equation is (x –
2
1) = −12(y − 5). Use a table of values to graph the
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parabola.
Because the parabola opens to the right, the vertex
is (11 − p , 4). Use the standard form of the
equation of a horizontal parabola and the point (20,
16) to find the equation.
4p (x – h) = (y – k)2
4p [20 – (11 − p )] = (16 – 4)2
4p (9 + p ) = 144
2
4p + 36p = 144
2
p + 9p − 36 = 0
(p + 12)(p − 3) = 0
p = −12 or 3
Because the parabola opens to the right, the value
of p must be positive. Therefore, p = 3. The vertex
is (8, 4), and the standard form of the equation is 12
2
(x – 8) = (y − 4) . Use a table of values to graph
the parabola.
Page 18
7-1 Parabolas
38. F(11, 4); opens right; contains (20, 16)
39. F(−4, 0); opens down; contains (4, −15)
SOLUTION:
SOLUTION:
Because the parabola opens to the right, the vertex
is (11 − p , 4). Use the standard form of the
equation of a horizontal parabola and the point (20,
16) to find the equation.
4p (x – h) = (y – k)2
Because the parabola opens down, the vertex is
(−4, −p ). Use the standard form of the equation of
a vertical parabola and the point (4, −15) to find the
equation.
4p (y – k) = (x – h)2
4p [−15 – (−p )] = [4 – (−4)]2
4p (−15 + p ) = 64
2
4p − 60p = 64
4p [20 – (11 − p )] = (16 – 4)2
4p (9 + p ) = 144
2
4p + 36p = 144
2
p + 9p − 36 = 0
(p + 12)(p − 3) = 0
p = −12 or 3
Because the parabola opens to the right, the value
of p must be positive. Therefore, p = 3. The vertex
is (8, 4), and the standard form of the equation is 12
2
(x – 8) = (y − 4) . Use a table of values to graph
the parabola.
2
p − 15p − 16 = 0
(p − 16)(p + 1) = 0
p = 16 or −1
Because the parabola opens down, the value of p
must be negative. Therefore, p = −1. The vertex is
(−4, 1), and the standard form of the equation is (x
2
+ 4) = −4(y − 1). Use a table of values to graph
the parabola.
40. F(1, 3); opens left; contains (−14, 11)
39. F(−4, 0); opens down; contains (4, −15)
SOLUTION:
Because the parabola opens down, the vertex is
(−4, −p ). Use the standard form of the equation of
a vertical parabola and the point (4, −15) to find the
equation.
4p (y – k) = (x – h)2
4p [−15 – (−p )] = [4 – (−4)]2
4p (−15 + p ) = 64
2
4p − 60p = 64
2
p − 15p − 16 = 0
(p − 16)(p + 1) = 0
p = 16 or −1
Because the parabola opens down, the value of p
must be negative. Therefore, p = −1. The vertex is
(−4, 1), and the standard form of the equation is (x
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2
+ 4) = −4(y − 1). Use a table of values to graph
the parabola.
SOLUTION:
Because the parabola opens to the left, the vertex
is (1 − p , 3). Use the standard form of the equation
of a horizontal parabola and the point (−14, 11) to
find the equation.
4p (x – h) = (y – k)2
4p [−14 – (1 − p )] = (11 – 3)2
4p (−15 + p ) = 64
2
4p − 60p = 64
2
p − 15p − 16 = 0
(p − 16)(p + 1) = 0
p = −1 or 16
Because the parabola opens to the left, the value of
p must be negative. Therefore, p = −1. The vertex
is (2, 3), and the standard form of the equation is −4
2
(x – 2) = (y − 3) . Use a table of values to graph
Page 19
the parabola.
7-1 Parabolas
40. F(1, 3); opens left; contains (−14, 11)
41. F(−5, −9); opens right; contains (10, −1)
SOLUTION:
SOLUTION:
Because the parabola opens to the left, the vertex
is (1 − p , 3). Use the standard form of the equation
of a horizontal parabola and the point (−14, 11) to
find the equation.
4p (x – h) = (y – k)2
Because the parabola opens to the right, the vertex
is (−5 − p , −9). Use the standard form of the
equation of a horizontal parabola and the point (10,
−1) to find the equation.
4p (x – h) = (y – k)2
4p [10 – (−5 − p )] = [−1 – (−9)]2
4p (15 + p ) = 64
2
4p + 60p = 64
4p [−14 – (1 − p )] = (11 – 3)2
4p (−15 + p ) = 64
2
4p − 60p = 64
2
p − 15p − 16 = 0
(p − 16)(p + 1) = 0
p = −1 or 16
Because the parabola opens to the left, the value of
p must be negative. Therefore, p = −1. The vertex
is (2, 3), and the standard form of the equation is −4
2
(x – 2) = (y − 3) . Use a table of values to graph
the parabola.
41. F(−5, −9); opens right; contains (10, −1)
2
p + 15p − 16 = 0
(p + 16)(p − 1) = 0
p = 1 or −16
Because the parabola opens to the right, the value
of p must be positive. Therefore, p = 1. The vertex
is (−6, −9), and the standard form of the equation is
2
4(x + 6) = (y + 9) . Use a table of values to graph
the parabola.
42. F(−7, 6); opens left; contains (−4, 10)
SOLUTION:
SOLUTION:
Because the parabola opens to the right, the vertex
is (−5 − p , −9). Use the standard form of the
equation of a horizontal parabola and the point (10,
−1) to find the equation.
4p (x – h) = (y – k)2
4p [10 – (−5 − p )] = [−1 – (−9)]2
4p (15 + p ) = 64
2
4p + 60p = 64
Because the parabola opens to the left, the vertex
is (−7 − p , 6). Use the standard form of the
equation of a horizontal parabola and the point (−4,
10) to find the equation.
4p (x – h) = (y – k)2
2
p + 15p − 16 = 0
(p + 16)(p − 1) = 0
p = 1 or −16
Because the parabola opens to the right, the value
of p must be positive. Therefore, p = 1. The vertex
is (−6, −9), and the standard form of the equation is
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2
4(x + 6) = (y + 9) . Use a table of values to graph
the parabola.
4p [−4 – (−7 − p )] = (10 – 6)2
4p (3 + p ) = 16
2
4p + 12p = 16
2
p + 3p − 4 = 0
(p + 4)(p − 1) = 0
p = −4 or 1
Because the parabola opens to the left, the value of
p must be negative. Therefore, p = −4. The vertex
is (−3, 6), and the standard form of the equation
is 20
Page
2
−16(x + 3) = (y − 6) . Use a table of values to
graph the parabola.
7-1 Parabolas
42. F(−7, 6); opens left; contains (−4, 10)
43. F(−5, −2); opens up; contains (−13, −2)
SOLUTION:
SOLUTION:
Because the parabola opens to the left, the vertex
is (−7 − p , 6). Use the standard form of the
equation of a horizontal parabola and the point (−4,
10) to find the equation.
4p (x – h) = (y – k)2
Because the parabola opens up, the vertex is (−5,
−2 − p ). Use the standard form of the equation of a
horizontal parabola and the point (−13, −2) to find
the equation.
4p (y – k) = (x – h)2
4p [−2 – (−2 − p )] = [−13 – (−5)]2
4p (p ) = 64
2
p = 16
p = ±4
Because the parabola opens up, the value of p must
be positive. Therefore, p = 4. The vertex is (−5,
−6), and the standard form of the equation is (x +
4p [−4 – (−7 − p )] = (10 – 6)2
4p (3 + p ) = 16
2
4p + 12p = 16
2
p + 3p − 4 = 0
(p + 4)(p − 1) = 0
p = −4 or 1
Because the parabola opens to the left, the value of
p must be negative. Therefore, p = −4. The vertex
is (−3, 6), and the standard form of the equation is
2
5) = 16(y + 6). Use a table of values to graph the
parabola.
2
−16(x + 3) = (y − 6) . Use a table of values to
graph the parabola.
44. ARCHITECTURE The entrance to an open–air
43. F(−5, −2); opens up; contains (−13, −2)
SOLUTION:
Because the parabola opens up, the vertex is (−5,
−2 − p ). Use the standard form of the equation of a
horizontal parabola and the point (−13, −2) to find
the equation.
4p (y – k) = (x – h)2
4p [−2 – (−2 − p )] = [−13 – (−5)]2
4p (p ) = 64
2
p = 16
p = ±4
Because the parabola opens up, the value of p must
be positive. Therefore, p = 4. The vertex is (−5,
−6), and the standard form of the equation is (x +
2
5) = 16(y + 6). Use a table of values to graph the
parabola.
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plaza has a parabolic arch above two columns. The
light in the center is located at the focus of the
parabola.
a. Write an equation that models the parabola.
b. Graph the equation.
SOLUTION:
a. Assume the ground represents the x–axis and
the left column the y–axis. Since the center of the
entrance is at 29 feet, the vertex for the parabolic
arch is located at the point (29, 28) and the focus is
located at the point (29, 14.9). Because the graph is
vertical, the focus is (h, k + p ), so the value of p is
Page 21
14.9 − (28) or −13.1.
Write the equation for the parabola in standard
7-1 Parabolas
Write an equation for the line tangent to each
parabola at each given point.
44. ARCHITECTURE The entrance to an open–air
plaza has a parabolic arch above two columns. The
light in the center is located at the focus of the
parabola.
a. Write an equation that models the parabola.
b. Graph the equation.
45.
SOLUTION:
The graph opens vertically. Determine the vertex
is written in
and focus.
. The
standard form. Because
vertex is (−7, 3) and the focus is (−7, 2.875). We
need to determine d, the distance between the
focus and the point of tangency, C.
SOLUTION:
a. Assume the ground represents the x–axis and
the left column the y–axis. Since the center of the
entrance is at 29 feet, the vertex for the parabolic
arch is located at the point (29, 28) and the focus is
located at the point (29, 14.9). Because the graph is
vertical, the focus is (h, k + p ), so the value of p is
14.9 − (28) or −13.1.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−13.1)(y – 28) = (x – 29)2
−52.4(y − 28) = (x − 29)2
The standard form of the equation is −52.4(y – 28)
2
= (x – 29) .
Graph the vertex and focus. Then make a table of
values to graph the parabola.
This is one leg of the isosceles triangle.
Use d to find A, the endpoint of the other leg of the
isosceles triangle. Since p is negative, the parabola
opens down and A will be above the focus.
Points A and C both lie on the line tangent to the
parabola. Find an equation of this line.
b. Use a table of values to graph the parabola.
46. y 2 =
Write an equation for the line tangent to each
parabola at each given point.
(x – 4), (24, 2)
SOLUTION:
The graph opens horizontally. Determine the vertex
2
45.
and focus. y =
SOLUTION:
form. Because 4p =
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The graph opens vertically. Determine the vertex
and focus.
(x − 4) is written in standard
is written in
,p =
. The vertex is
(4,
Page 22
0) and the focus is (4.05, 0). We need to determine
d, the distance between the focus and the point of
y–2 =
y =
7-1 Parabolas
46. y 2 =
−
x+
47. (x + 6)2 = 3(y – 2), (0, 14)
(x – 4), (24, 2)
SOLUTION:
SOLUTION:
The graph opens vertically. Determine the vertex
The graph opens horizontally. Determine the vertex
and focus. (x + 6) = 3(y − 2) is written in standard
2
and focus. y =
(x − 4) is written in standard
form. Because 4p =
,p =
. The vertex is (4,
0) and the focus is (4.05, 0). We need to determine
d, the distance between the focus and the point of
tangency, C.
This is one leg of the isosceles triangle.
d =
2
form. Because 4p = 3, p =
. The vertex is (−6, 2)
and the focus is (−6, 2.75). We need to determine
d, the distance between the focus and the point of
tangency, C.
This is one leg of the isosceles triangle.
d =
=
=
= 12.75
= 20.05
Use d to find A, the endpoint of the other leg of the
isosceles triangle. Since p is positive, the parabola
opens to the right and A will be to the left of the
focus.
A = (4.05 – 20.05, 0) or (−16, 0)
Points A and C both lie on the line tangent to the
parabola. Find an equation of this line.
m
or
=
(x – 24)
y–2 =
−
Points A and C both lie on the line tangent to the
parabola. Find an equation of this line.
or 4
=
m
y – y 1 = m(x – x1)
y – 14 = 4(x – 0)
y – y 1 = m(x – x1)
y–2 =
Use d to find A, the endpoint of the other leg of the
isosceles triangle. Since p is positive, the parabola
opens up and A will be below the focus.
A = (−6, 2.75 − 12.75) or (−6, −10)
y − 14 = 4x
y = 4x + 14
48. (x – 3)2 = y + 4, (−1, 12)
y =
x+
SOLUTION:
The graph opens vertically. Determine the vertex
2
and focus. (x − 3) = y + 4 is written in standard
47. (x + 6)2 = 3(y – 2), (0, 14)
form. Because 4p = 1, p =
SOLUTION:
The graph opens vertically. Determine the vertex
2
and focus. (x + 6) = 3(y − 2) is written in standard
form. Because 4p = 3, p =
. The vertex is (3, −4)
and the focus is (3, −3.75). We need to determine
d, the distance between the focus and the point of
tangency, C.
. The vertex is (−6, 2)
and the focus is (−6, 2.75). We need to determine
d, the distance between the focus and the point of
tangency, C.
This is one leg of the isosceles triangle.
d =
=
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This is one leg of the isosceles triangle.
d =
Page 23
= 16.25
y – 12 = −8[x – (−1)]
y – 14 = 4(x – 0)
7-1
y − 12 = −8x − 8
y = −8x + 4
y − 14 = 4x
y = 4x + 14
Parabolas
48. (x – 3)2 = y + 4, (−1, 12)
49. −0.25(x – 6)2 = y – 9, (10, 5)
SOLUTION:
SOLUTION:
The graph opens vertically. Determine the vertex
2
and focus. (x − 3) = y + 4 is written in standard
form. Because 4p = 1, p =
. The vertex is (3, −4)
and the focus is (3, −3.75). We need to determine
d, the distance between the focus and the point of
tangency, C.
This is one leg of the isosceles triangle.
d =
=
The graph opens vertically. Determine the vertex
and focus.
2
−0.25(x − 6) = y − 9
2
(x – 6) = −4(y – 9)
Because 4p = −4, p = −1. The vertex is (6, 9) and
the focus is (6, 8). We need to determine d, the
distance between the focus and the point of
tangency, C.
This is one leg of the isosceles triangle.
d =
=
= 16.25
=5
Use d to find A, the endpoint of the other leg of the
isosceles triangle. Since p is positive, the parabola
opens up and A will be below the focus.
A = (3, −3.75 − 16.25) or (3, −20)
Use d to find A, the endpoint of the other leg of the
isosceles triangle. Since p is negative, the parabola
opens down and A will be above the focus.
A = (6, 8 + 5) or (6, 13)
Points A and C both lie on the line tangent to the
parabola. Find an equation of this line.
Points A and C both lie on the line tangent to the
parabola. Find an equation of this line.
m
=
or −8
y – y 1 = m(x – x1)
m
=
or −2
y – y 1 = m(x – x1)
y – 12 = −8[x – (−1)]
y – 5 = −2(x – 10)
y − 12 = −8x − 8
y = −8x + 4
y − 5 = −2x + 20
y = −2x + 25
49. −0.25(x – 6)2 = y – 9, (10, 5)
50. −4x = (y + 5)2, (0, −5)
SOLUTION:
SOLUTION:
The graph opens vertically. Determine the vertex
and focus.
2
−0.25(x − 6) = y − 9
The graph opens horizontally. Determine the vertex
2
(x – 6) = −4(y – 9)
Because 4p = −4, p = −1. The vertex is (6, 9) and
the focus is (6, 8). We need to determine d, the
distance between the focus and the point of
tangency, C.
This is one leg of the isosceles triangle.
d =
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2
and focus. −4x = (y + 5) is written in standard
form. Because 4p = −4, p = −1. The vertex is (0,
−5) and the focus is (−1, −5). We need to
determine d, the distance between the focus and
the point of tangency, C.
This is one leg of the isosceles triangle.
d =
=
=1
Page 24
=5
Use d to find A, the endpoint of the other leg of the
SOLUTION:
y – 5 = −2(x – 10)
7-1
Since the directrix is y = 4, the graph opens
vertically. Since p is negative, the graph will open
down.
y − 5 = −2x + 20
y = −2x + 25
Parabolas
50. −4x = (y + 5)2, (0, −5)
52. y 2 = −8(x – 6)
SOLUTION:
SOLUTION:
The graph opens horizontally. Determine the vertex
Because the y–term is squared, the parabola opens
horizontally. Since 4p = −8, p is −2. Since p is
negative, the graph will open towards the left.
2
and focus. −4x = (y + 5) is written in standard
form. Because 4p = −4, p = −1. The vertex is (0,
−5) and the focus is (−1, −5). We need to
determine d, the distance between the focus and
the point of tangency, C.
53. vertex (−5, 1), focus (−5, 3)
SOLUTION:
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k
+ p ), so the value of p is 3 − 1or 2. Because p is
positive, the graph opens up.
This is one leg of the isosceles triangle.
d =
=
54. focus (7, 10), directrix x = 1
=1
SOLUTION:
Since the directrix is x = 1, the graph opens
horizontally. Since the directrix is to the left of the
focus, the graph opens to the right.
Use d to find A, the endpoint of the other leg of the
isosceles triangle. Since p is negative, the parabola
opens to the left and A will be to the right of the
focus.
A = (−1 + 1, −5) or (0, −5)
Write an equation for each parabola.
Points A and C are the same point. Attempting to
find the slope of a line using identical points results
in an undefined slope.
m =
or undefined
55.
Since the parabola opens horizontally and the slope
of the tangent line is undefined, the line tangent to
the graph at the vertex will be a vertical line. Since
the x–coordinate of the vertex is 0, the equation for
the line tangent to the parabola through this point is
x = 0.
SOLUTION:
The focus and vertex share the same x–coordinate.
The focus is (h, k + p ), so the value of p is 4 − 5 or
−1.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−1)(y – 5) = (x – 3)2
Determine the orientation of each parabola.
51. directrix y = 4, p = −2
SOLUTION:
−4(y − 5) = (x − 3)2
Since the directrix is y = 4, the graph opens
vertically. Since p is negative, the graph will open
down.
52. y 2 = −8(x – 6)
SOLUTION:
Because the y–term is squared, the parabola opens
horizontally. Since 4p = −8, p is −2. Since p is
negative, the graph will open towards the left.
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53. vertex (−5, 1), focus (−5, 3)
SOLUTION:
56.
SOLUTION:
Page 25
The vertex is (0, −4) and the directrix is x = −2.
Since the directrix is x = h − p or −2 = 0 − p , p is 2.
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−1)(y – 5) = (x – 3)2
7-1 Parabolas
−4(y − 5)
4p (x – h) = (y – k)2
4(−4)[x – (−5)] = (y – 1)2
−16(x + 5) = (y − 1)2
2
= (x − 3)
56.
58.
SOLUTION:
SOLUTION:
The vertex is (0, −4) and the directrix is x = −2.
Since the directrix is x = h − p or −2 = 0 − p , p is 2.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(2)(x – 0) = [y – (−4)]2
8x = (y + 4)2
The focus is (2, −2) and the directrix is y = k − p =
−12. The y–coordinate of the focus is k + p = −2.
Solve the system of equations.
k − p = −12
k + p = −2
2k = −14
k = −7
Substitute k = −7 into the equation for the directrix
to find that p = 5.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(5)[y – (−7)] = (x – 2)2
20(y + 7) = (x − 2)2
57.
SOLUTION:
The focus is (−9, 1) and the directrix is x = −1. The
directrix is x = h − p or h − p = −1 and the x–
coordinate of the focus is h + p = −9.
Solve the system of equations.
h − p = −1
h + p = −9
2h = −10
h = −5
Substitute h = −5 into the equation for the directrix
to find that p is –4.
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−4)[x – (−5)] = (y – 1)2
−16(x + 5) = (y − 1)2
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59. BRIDGES The arch of the railroad track bridge
below is in the shape of a parabola. The two main
support towers are 208 meters apart and 80 meters
tall. The distance from the top of the parabola to
the water below is 60 meters.
a. Write an equation that models the shape of the
arch. Let the railroad track represent the x–axis.
b. Two vertical supports attached to the arch are
equidistant from the center of the parabola as
shown in the diagram. Find their lengths if they are
86.4 meters apart.
SOLUTION:
a. If the railroad track represents the x–axis and
we assume that the vertex lies on the y–axis, then
the vertex is at the point (0, −20). The arch meets
each support tower 104 meters to the left and to
the right of the vertex and 80 feet below the
railroad or the x–axis. Thus, two points on the
Page
parabola are at (−104, −80) and (104, −80). Find
p 26
using the vertex and a point on the parabola.
4p (y – k) = (x – h)2
7-1
a. If the railroad track represents the x–axis and
we assume that the vertex lies on the y–axis, then
the vertex is at the point (0, −20). The arch meets
each support tower 104 meters to the left and to
Parabolas
the right of the vertex and 80 feet below the
railroad or the x–axis. Thus, two points on the
parabola are at (−104, −80) and (104, −80). Find p
using the vertex and a point on the parabola.
4p (y – k) = (x – h)2
4p [−80 – (−20)] = (104 – 0)2
−240p = 10816
p = −45.066
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−45.066)[y – (−20)] = (x – 0)2
−180.27(y + 20) = x2
b. If the supports are 86.4 meters apart and are
equidistant from the center of the parabola, then
they are located 43.2 meters from the center of the
parabola. Therefore, the supports meet the
parabola when x = 43.2. Find y when x = 43.2.
−180.27(y + 20) = x2
−180.27(y + 20) = 43.22
−180.27(y + 20) = 1866.24
y + 20 = −10.35
y = −30.35
y = −30.35
Therefore, the supports meets the parabola at the
points (43.2, −30.35) and (−43.2, −30.35) and the
supports are 30.35 meters in length.
Write an equation for and graph a parabola
with each set of characteristics.
60. vertex (1, 8), contains (11, 13), opens vertically
SOLUTION:
If the graph opens vertically, then the standard
2
form is 4p (y − k) = (x − h) . Find p using the
vertex and the point on the parabola.
4p (y – k) = (x – h)2
4p (13 – 8) = (11 – 1)2
20p = 100
p =5
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(5)(y – 8) = (x – 1)2
20(y − 8) = (x − 1)2
Therefore, the supports meets the parabola at the
points (43.2, −30.35) and (−43.2, −30.35) and the
supports are 30.35 meters in length.
Write an equation for and graph a parabola
with each set of characteristics.
60. vertex (1, 8), contains (11, 13), opens vertically
61. vertex (−6, 4), contains (−10, 8), opens horizontally
SOLUTION:
SOLUTION:
If the graph opens vertically, then the standard
2
form is 4p (y − k) = (x − h) . Find p using the
vertex and the point on the parabola.
4p (y – k) = (x – h)2
4p (13 – 8) = (11 – 1)2
20p = 100
p =5
If the graph opens horizontally, then the standard
2
form is 4p (x − h) = (y − k) . Find p using the
vertex and the point on the parabola.
4p (x – h) = (y – k)2
4p [−10 – (−6)] = (8 – 4)2
−16p = 16
p = −1
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(5)(y – 8) = (x – 1)2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−1)[x – (−6)] = (y – 4)2
20(y − 8) = (x − 1)2
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−4(x + 6) = (y − 4)2
Page 27
4p (−2 – k) = (0 – h)2
−8p − 4k p = h 2
7-1 Parabolas
61. vertex (−6, 4), contains (−10, 8), opens horizontally
SOLUTION:
If the graph opens horizontally, then the standard
2
form is 4p (x − h) = (y − k) . Find p using the
vertex and the point on the parabola.
4p (x – h) = (y – k)2
4p [−10 – (−6)] = (8 – 4)2
−16p = 16
p = −1
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(−1)[x – (−6)] = (y – 4)2
Solve this system of equations. Subtract Equation 2
from Equation 1.
−56p − 4k p = 144 + 24h + h 2
−8p − 4k p = h 2
−48p = 144 + 24h
Subtract Equation 2 from Equation 3.
−20p − 4k p = 36 − 12h + h 2
−8p − 4k p = h 2
−12p = 36 − 12h
Solve the system consisting of the two new
equations.
−48p = 144 + 24h
−12p = 36 − 12h
−48p = 144 + 24h
−48p = 144 − 48h
0 = −24h
0 =h
−4(x + 6) = (y − 4)2
62. opens vertically, passes through points (−12, −14),
(0, −2), and (6, −5)
SOLUTION:
If the graph opens vertically, then the standard
2
form is 4p (y − k) = (x − h) . Find an equation for
the parabola for each set of points.
Equation 1:
4p (y – k) = (x – h)2
4p (−14 – k) = (−12 – h)2
−56p − 4k p = 144 + 24h + h 2
Equation 2:
4p (y – k) = (x – h)2
4p (−2 – k) = (0 – h)2
−8p − 4k p = h 2
Equation 3:
4p (y – k) = (x – h)2
2
(−5 – k)
= (6 – h)
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−20p − 4k p = 36 − 12h + h
Equation 3:
4p (y – k) = (x – h)2
4p (−5 – k) = (6 – h)2
−20p − 4k p = 36 − 12h + h 2
Substitute h = 0 into one of the equations and solve
for p .
−12p = 36 − 12h
−12p = 36 − 12(0)
−12p = 36
p = −3
Substitute p = –3 and h = 0 into one of the original
equations and solve for k.
−8p − 4k p = h 2
−8(−3) − 4(−3)k = 02
24 + 12k = 0
12k = −24
k = −2
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−3)[y – (−2)] = (x – 0)2
−12(y + 2) = x2
Page 28
2
7-1
form using the values of h, p , and k.
4p (y – k) = (x – h)2
4(−3)[y – (−2)] = (x – 0)2
Parabolas
−12(y + 2) = x2
Solve the system consisting of the two new
equations.
−24p = −8 + 8k
40p = 40 − 8k
16p = 32
p =2
Substitute p = 2 into one of the equations and solve
for k.
40p = 40 − 8k
40(2) = 40 − 8k
80 = 40 − 8k
40 = −8k
−5 = k
63. opens horizontally, passes through points (−1, −1),
(5, 3), and (15, 7)
SOLUTION:
If the graph opens horizontally, then the standard
2
form is 4p (x − h) = (y − k) . Find an equation for
the parabola for each set of points.
Equation 1:
4p (x – h) = (y – k)2
4p (−1 – h) = (−1 – k)2
−4p − 4hp = 1 + 2k + k 2
Equation 2:
4p (x – h) = (y – k)2
4p (5 – h) = (3 – k)2
20p − 4hp = 9 − 6k + k 2
Equation 3:
4p (x – h) = (y – k)2
4p (15 – h) = (7 – k)2
Substitute p = 2 and k = −5 into one of the original
equations and solve for h.
−4p − 4hp = 1 + 2k + k 2
−4(2) − 4(2)h = 1 + 2(−5) + (−5)2
−8 − 8h = 16
−8h = 24
h = −3
Write the equation for the parabola in standard
form using the values of h, p , and k.
4p (x – h) = (y – k)2
4(2)[x – (−3)] = [y – (−5)]2
8(x + 3) = (y + 5)2
60p − 4hp = 49 − 14k + k 2
Solve this system of equations. Subtract Equation 2
from Equation 1.
−4p − 4hp = 1 + 2k + k 2
20p − 4hp = 9 − 6k + k 2
−24p = −8 + 8k
Subtract Equation 2 from Equation 3.
60p − 4hp = 49 − 14k + k 2
20p − 4hp = 9 − 6k + k 2
40p = 40 − 8k
Solve the system consisting of the two new
equations.
−24p = −8 + 8k
40p = 40 − 8k
16p = 32
p by
= 2Cognero
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Substitute p = 2 into one of the equations and solve
64. SOUND Parabolic reflectors with microphones
located at the focus are used to capture sounds
from a distance. The sound waves enter the
reflector and are concentrated toward the
microphone.
a. How far from the reflector should a microphone
be placed if the reflector has a width of 3 feetPage
and29
a depth of 1.25 feet?
b. Write an equation to model a different parabolic
reflector that is 4 feet wide and 2 feet deep, if the
4p (1 − 3) = (7 − 5)2
−8p = 4
p = −0.5
7-1 Parabolas
a. How far from the reflector should a microphone
be placed if the reflector has a width of 3 feet and
a depth of 1.25 feet?
b. Write an equation to model a different parabolic
reflector that is 4 feet wide and 2 feet deep, if the
vertex of the reflector is located at (3, 5) and the
parabola opens to the left.
c. Graph the equation. Specify the domain and
range.
SOLUTION:
a. The vertex and focus of a parabola in standard
form are (h, k) and (h + p , k) respectively. Find p
to determine the distance the microphone should be
from the reflector. Sketch a coordinate plane on top
of the reflector with the vertex located at the origin.
The width represents the distance the reflector
spans vertically. Since the reflector has a width of
3 feet, 1.5 feet of the reflector will exist on each
side of the x–axis. The depth represents the
distance the reflector spans horizontally along the
x–axis. Using a depth of 1.25 feet and a width of
1.5 feet, the points (1.25, 1.5) and (1.25, −1.5) must
lie on the reflector. Since the parabola opens to the
2
right, the standard form is 4p (x − h) = (y − k) .
Substitute the values of x, y, h, and k to solve for p .
4p (x − h) = (y − k)2
4p (1.25 − 0) = (1.5 − 0)2
5p = 2.25
p = 0.45
The focus of the parabola is at (h + p , k) or (0.45,
0). Thus, the microphone should be placed at the
point (0.45, 0). This is 0.45 feet from the vertex of
the reflector.
b. Since the parabola opens to the left, the standard
2
form is 4p (x − h) = (y − k) . The vertex is (3, 5).
The depth represents the distance the reflector
spans horizontally and the width represents the
distance the reflector spans vertically. Points that
are 2 feet to the left of the vertex and 2 feet above
and below it will lie on the reflector. Thus, the
points (1, 7) and (1, 3) lie on the parabola.
Substitute the values of x, y, h, and k to solve for p .
4p (x − h) = (y − k)2
4p (1 − 3) = (7 − 5)2
−8p = 4
p = −0.5
Substitute values for h, k, and p to write an
equation
modelbythe
reflector.
eSolutions
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4p (x − h) = (y − k)2
4(–0.5)(x − = (y − 5)2
Substitute values for h, k, and p to write an
equation to model the reflector.
4p (x − h) = (y − k)2
4(–0.5)(x − = (y − 5)2
3)
−2(x − 3) = (y − 5)2
c. Use a table of values to graph the parabola.
Since the reflector has a depth of 2 feet, graph the
parabola for 1 ≤ x ≤ 3.
Determine the point of tangency for each
equation and line.
65. (x + 2)2 = 2y
y = 4x
SOLUTION:
The point of tangency occurs at the solution for the
system of equations.
Use substitution to solve the system.
2
(x + 2) = 2y
2
= 2(4x)
2
x + 4x + 4 = 8x
2
x − 4x + 4 = 0
2
(x − 2) = 0
(x + 2)
The solution to the system occurs when x = 2. To
find the y–coordinate, substitute x = 2 in one of the
original equations.
y = 4x
y = 4(2)
y =8
The point of tangency occurs at (2, 8).
66. (y – 8)2 = 12x
y = x + 11
SOLUTION:
Page 30
The point of tangency occurs at the solution for the
system of equations.
y = x + 11
y = 3 + 11
y = 14
y = 4(2)
y =8
7-1 Parabolas
The point of tangency occurs at (2, 8).
66. (y – 8)2 = 12x
The point of tangency occurs at (3, 14).
67. (y + 3)2 = −x + 4
y = x + 11
SOLUTION:
The point of tangency occurs at the solution for the
system of equations.
Use substitution to solve the system.
2
(y − 8) = 12x
[(x + 11) − 8] = 12x
SOLUTION:
The point of tangency occurs at the solution for the
system of equations.
Use substitution to solve the system.
2
(y + 3) = −x + 4
2
= −x + 4
2
(x + 3) = 12x
2
x + 6x + 9 = 12x
2
x − 6x + 9 = 0
2
(x − 3)
= −x + 4
=0
= −x + 4
The solution to the system occurs when x = 3. To
find the y–coordinate, substitute x = 3 into one of
the original equations.
y = x + 11
y = 3 + 11
y = 14
=0
The solution to the system occurs when x = 0. To
find the y–coordinate, substitute x = 0 into one of
the original equations.
The point of tangency occurs at (3, 14).
67. (y + 3)2 = −x + 4
y = −1
The point of tangency occurs at (0, −1).
68. (x + 5)2 = −4(y + 1)
SOLUTION:
The point of tangency occurs at the solution for the
system of equations.
Use substitution to solve the system.
2
(y + 3) = −x + 4
= −x + 4
= −x + 4
y = 2x + 13
SOLUTION:
The point of tangency occurs at the solution for the
system of equations.
Use substitution to solve the system.
2
(x + 5) = −4(y + 1)
2
(x + 5) = −4[(2x + 13) + 1]
2
= −4(2x + 14)
x + 10x + 25 = −8x − 56
2
x + 18x + 81 = 0
2
(x + 9) = 0
(x + 5)
2
= −x + 4
=0
The solution to the system occurs when x = 0. To
find the y–coordinate, substitute x = 0 into one of
the original equations.
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y = −1
The point of tangency occurs at (0, −1).
The solution to the system occurs when x = −9. To
find the y–coordinate, substitute x = −9 into one of
the original equations.
y = 2x + 13
y = 2(−9) + 13
y = −5
Page 31
The point of tangency occurs at (−9, −5).
SOLUTION:
y = −1
7-1 Parabolas
The point of tangency occurs at (0, −1).
68. (x + 5)2 = −4(y + 1)
y = 2x + 13
SOLUTION:
The point of tangency occurs at the solution for the
system of equations.
Use substitution to solve the system.
2
(x + 5) = −4(y + 1)
2
(x + 5) = −4[(2x + 13) + 1]
2
(x + 5) = −4(2x + 14)
2
x + 10x + 25 = −8x − 56
2
x + 18x + 81 = 0
2
(x + 9) = 0
The solution to the system occurs when x = −9. To
find the y–coordinate, substitute x = −9 into one of
the original equations.
y = 2x + 13
y = 2(−9) + 13
y = −5
The point of tangency occurs at (−9, −5).
69. ILLUMINATION In a searchlight, the bulb is
placed at the focus of a parabolic mirror 1.5 feet
from the vertex. This causes the light rays from the
bulb to bounce off the mirror as parallel rays, thus
providing a concentrated beam of light.
a. Sketch a coordinate plane on top of the
searchlight with the vertex at the origin. The focus
is 1.5 feet from the vertex located at the point (1.5,
0). Since the focal diameter of the bulb is 2 feet, 1
foot will lie above and below the x–axis. Therefore,
the points (1.5, 1) and (1.5, −1) lie on the parabola.
Substitute the values of x, y, h, and k to solve for p .
4p (x − h) = (y − k)2
4p (1.5 − 0) = (1 − 0)2
6p = 1
p
=
Substitute values for h, k, and p to write an
equation to model the reflector.
4p (x − h) = (y − k)2
4
(x − 0) = (y − 0)2
x
=y
2
b. Write an equation in standard form for a
parabola with a focal diameter of 3 feet. Sketch a
coordinate plane on top of the new searchlight with
the vertex at the origin. The focus is still 1.5 feet
from the vertex located at the point (1.5, 0). Since
the focal diameter of the bulb is 3 feet, 1.5 feet will
lie above and below the x–axis. Therefore, the
points (1.5, 1.5) and (1.5, –1.5) lie on the parabola.
Substitute the values of x, y, h, and k to solve for p .
4p (x − h) = (y − k)2
4p (1.5 − 0) = (1.5 − 0)2
6p = 2.25
p = 0.375
Substitute values for h, k, and p to write an
equation to model the reflector.
4p (x − h) = (y − k)2
a.Write an equation for the parabola if the focal
diameter of the bulb is 2 feet, as shown in the
diagram.
b.Suppose the focal diameter is increased to 3 feet.
If the depth of both searchlights is 3.5 feet, how
much greater is the width of the opening of the
larger light? Round to the nearest hundredth.
4(0.375)(x − 0) = (y − 0)2
1.5x = y 2
The depth of both searchlights is 3.5 feet. The
depth represents the distance the reflector spans
horizontally and the width represents the distance
the reflector spans vertically. To find the width of
the first searchlight, substitute x = 3.5 into the
equation found in part a and solve for y.
SOLUTION:
a. Sketch a coordinate plane on top of the
searchlight with the vertex at the origin. The focus
is 1.5 feet from the vertex located at the point (1.5,
0). Since the focal diameter of the bulb is 2 feet, 1
foot will lie above and below the x–axis. Therefore,
eSolutions
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theManual
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(1.5, 1) by
and
(1.5, −1) lie on the parabola.
Substitute the values of x, y, h, and k to solve for p .
4p (x − h) = (y − k)2
x
=y
2
(3.5) = y 2
=y
2
Page 32
±
=y
(3.5) = y 2
±
or 2.29 = y
At a depth of 3.5 feet, the reflector will span 2.29
feet above and below the x-axis for a total width of
2(2.29) or 4.58 feet. The difference between the
two widths is 4.58 – 3.06 or 1.52 feet.
2
70. PROOF Use the standard form of the equation of
equation found in part a and solve for y.
x
7-1 Parabolas
=y
=y
a parabola to prove the general form of the
equation of a parabola.
=y
±
±
2
SOLUTION:
The general form of equations for conic sections is
or ±1.53 = y
2
At a depth of 3.5 feet, the reflector will span 1.53
feet above and below the x-axis for a total width of
2(1.53) or 3.06 feet. Repeat the same process to
find the width of the second reflector.
1.5x = y 2
1.5(3.5) = y 2
5.25 = y 2
±
or 2.29 = y
At a depth of 3.5 feet, the reflector will span 2.29
feet above and below the x-axis for a total width of
2(2.29) or 4.58 feet. The difference between the
two widths is 4.58 – 3.06 or 1.52 feet.
70. PROOF Use the standard form of the equation of
a parabola to prove the general form of the
equation of a parabola.
2
Ax + Bxy + Cy + Dx + Ey + F = 0, where A, B,
C, D, E, and F are constants and A, B, and C
cannot all be zero. Expand each standard form of
equations for a parabola.
2
(y – k) = 4p (x – h)
2
y 2 – 2k y + k = 4px – 4ph
2
y 2 – 4px – 2k y + k = 0
+ 4ph
2
y + (−4p )x + (−2k)y = 0
2
+ (k + 4ph )
2
y + Dx + Ey + F = 0
Once values are given for p , k, and h, these
variables become constants. Thus, they may be
represented by constants D, E, and F. Repeat the
process using the other standard form of equation
for a parabola.
2
(x – h) = 4p (y – k)
2
SOLUTION:
The general form of equations for conic sections is
2
2
Ax + Bxy + Cy + Dx + Ey + F = 0, where A, B,
C, D, E, and F are constants and A, B, and C
cannot all be zero. Expand each standard form of
equations for a parabola.
2
(y – k) = 4p (x – h)
2
y 2 – 2k y + k = 4px – 4ph
2
y 2 – 4px – 2k y + k = 0
+ 4ph
2
y + (−4p )x + (−2k)y = 0
2
+ (k + 4ph )
2
y + Dx + Ey + F = 0
2
x – 2hx + h
2
2
x – 2hx – 4py + h + 4p k
2
= 4py − 4p k
=0
2
x + (−2h)x + (−4p )y + (h + = 0
4p k )
2
x + Dx + Ey + F = 0
71. The latus rectum of a parabola is the line segment
that passes through the focus, is perpendicular to
the axis of the parabola, and has endpoints on the
parabola. The length of the latus rectum is |4p |
units, where p is the distance from the vertex to the
focus.
Once values are given for p , k, and h, these
variables become constants. Thus, they may be
represented by constants D, E, and F. Repeat the
process using the other standard form of equation
for a parabola.
2
(x – h) = 4p (y – k)
2
2
x – 2hx + h
2
2
x – 2hx – 4py + h + 4p k
2
2
= 4py − 4p k
=0
x + (−2h)x + (−4p )y + (h + = 0
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4p k )
2
x + Dx + Ey + F = 0
a.Write an equation for the parabola with vertex at
(−3, 2), axis y = 2, and latus rectum 8 units long.
b.Prove that the endpoints of the latus rectum and
point of intersection of the axis and directrix are
the
Page
33
vertices of a right isosceles triangle.
SOLUTION:
isosceles right triangle. Therefore, XDF is
.
This also means that FDW is
. Resultantly,
XDW is a right angle. Thus, ∆XDW is an
isosceles right triangle.
a.Write an equation for the parabola with vertex at
7-1 Parabolas
(−3, 2), axis y = 2, and latus rectum 8 units long.
b.Prove that the endpoints of the latus rectum and
point of intersection of the axis and directrix are the
vertices of a right isosceles triangle.
72. SOLAR ENERGY A solar furnace in France’s
Eastern Pyrenees uses a parabolic mirror that is
illuminated by sunlight reflected off a field of
heliostats, which are devices that track and
redirect sunlight. Experiments in solar research are
performed in the focal zone part of a tower. If the
parabolic mirror is 6.25 meters deep, how many
meters in front of the parabola is the focal zone?
SOLUTION:
a. Since the parabola has an axis of y = 2, the
parabola will open horizontally and will have the
2
standard form of (y – k) = 4p (x – h). The latus
rectum is 8 units long. This can be used to solve for
p.
|4p | = 8, so p = ±2.
Substitute values for h, k, and p to write an equation
for the parabola.
4p (x − h) = (y − k)2
4(±2)[x − (−3)] = (y − 2)2
±8(x + 3) = (y − 2)2
SOLUTION:
The focal zone is located at the focus of the
parabolic mirror. The vertex and focus of a
parabola in standard form are (h, k) and (h + p , k)
respectively. Find p to determine the distance the
focal zone is from the mirror. Sketch a coordinate
plane on top of the diagram with the vertex located
at the origin. The width represents the distance the
reflector spans vertically and is shown to be 40
meters. Since the reflector has a width of 40
meters, 20 meters of the reflector will exist on each
side of the x–axis. The depth represents the
distance the reflector spans horizontally along the
x–axis. Using a depth of 6.25 meters and a width
of 20 meters, the points (6.25, 20) and (6.25, −20)
must lie on the mirror. Since the parabola opens to
b.
To prove that the endpoints of the latus rectum X
and W and the point of intersection of the axis and
directrix D are the vertices of a right isosceles
triangle ∆XDW, we need to show that XDW is a
right angle and that
. Since
,
, and
XFD
2
the right, the standard form is 4p (x − h) = (y − k) .
Substitute the values of x, y, h, and k to solve for p .
4p (x − h) = (y − k)2
4p (6.25 − 0) = (20 − 0)2
25p = 400
p = 16
WFD, ∆XFD
∆FWD by SAS. Thus,
. To prove that
XDW is a right angle, we can first prove that
XDF and FDW are both
angles. Since
and XFD is a right angle, ∆XFD is an
The focus of the parabola is at (h + p , k) or (16, 0).
Thus, the focal zone is at the point (16, 0). This is
16 meters from the vertex of the mirror.
isosceles right triangle. Therefore, XDF is
.
This also means that FDW is
. Resultantly,
XDW is a right angle. Thus, ∆XDW is an
isosceles right triangle.
Write a possible equation for a parabola with
focus F such that the line given is tangent to
the parabola.
72. SOLAR ENERGY A solar furnace in France’s
Eastern Pyrenees uses a parabolic mirror that is
illuminated by sunlight reflected off a field of
heliostats, which are devices that track and
redirect sunlight. Experiments in solar research are
performed
in the focal
zone part of a tower. If the
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parabolic mirror is 6.25 meters deep, how many
meters in front of the parabola is the focal zone?
Page 34
73.
The solutions to this equation are x = 0 and 4. Since
A has an x–coordinate of 0, the point of tangency
must have an x–coordinate of 4. Substitute x = 4 to
solve for y.
y = 2x − 2
y = 2(4) − 2
y =6
Thus, the focal zone is at the point (16, 0). This is
16 meters from the vertex of the mirror.
Write a possible equation for a parabola with
7-1 Parabolas
focus F such that the line given is tangent to
the parabola.
The point of tangency is located at (4, 6). Since the
parabola opens upward, the standard form of the
2
equation is (x − h) = 4p (y − k) and the focus is
located at (h, k + p ). The focus is at the point (0,
3); thus, h = 0 and k + p = 3 or k = 3 − p . Use, h,
the standard form of the equation, the point of
tangency, and the equation for k to solve for p .
2
(x − h) = 4p (y − k)
73.
SOLUTION:
Find point A, the point of intersection of the axis
and the line tangent to the parabola.
The equation for the tangent line is y = 2x − 2.
Since point A lies on the y–axis, the x–coordinate is
0. Substitute x = 0 to find y.
y = 2x − 2
y = 2(0) − 2
y = −2
2
(4 − 0)
16
16
16
= 4p (6 − k)
= 24p − 4p k
= 24p − 4p (3 − p )
= 24p − 12p + 4p
0 = 4p 2 + 12p − 16
0 = 4(p + 4)(p − 1)
2
The solutions to this equation are p = –4 and 1.
Since the parabola opens upward, p is positive;
thus, p = 1. Use p to find k. k = 3 − (1) or 2. The
vertex of the parabola is (0, 2). Use the vertex and
p to write the standard form of the equation.
2
(x − h) = 4p (y − k)
A is at (0, −2). The segment formed by FA
represents one of the legs of the isosceles triangle
formed by F, A, and the point of tangency. Find the
length of this leg.
2
(x − 0)
d =5
x
2
= 4(1)(y − 2)
= 4(y − 2)
The lengths of the legs of the isosceles triangle are
5 units. This is the distance from F to the point of
tangency (x, y). Use the distance formula and the
tangent line to solve for the point of tangency.
d
5
25 = x2 + (y − 3)2
25 = x2 + y 2 − 6y + 9
25 = x2 + (2x − 2)2 − 6(2x − 2) + 9
25 = x2 + 4x2 − 8x + 4 − 12x + 12 + 9
25 = 5x2 − 20x + 25
0 = 5x2 − 20x
0 = 5x(x − 4)
74.
SOLUTION:
Find point A, the point of intersection of the axis
and the line tangent to the parabola.
The equation for the tangent line is y =
The solutions to this equation are x = 0 and 4. Since
A has an x–coordinate of 0, the point of tangency
must have an x–coordinate of 4. Substitute x = 4 to
solve for y.
y = 2x − 2
y = 2(4) − 2
y =Manual
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The point of tangency is located at (4, 6). Since the
parabola opens upward, the standard form of the
x + 2.
Since point A lies on the x–axis, the y–coordinate is
0. Substitute y = 0 to find x.
y
=
x+2
0 =
x+2
Page 35
−2 =
x
The equation for the tangent line is y =
x + 2.
y=
Since point A lies on the x–axis, the y–coordinate is
0. Substitute y = 0 to find x.
y =4
7-1 Parabolas
y
=
x+2
0 =
x+2
(4) + 2
The point of tangency is located at (4, 4). Since the
parabola opens to the right, the standard form of
2
−2 =
the equation is 4p (x − h) = (y − k) and the focus is
located at (h + p , k). The focus is at the point (1,
0); thus, k = 0 and h + p = 1 or h = 1 − p . Use, k,
the standard form of the equation, the point of
tangency, and the equation for h to solve for p .
4p (x − h) = (y − k)2
4p (4 − h) = (4 − 0)2
16p − 4ph = 16
16p − 4p (1 − p ) = 16
2
16p − 4p + 4p = 16
2
4p + 12p − 16 = 0
x
−4 = x
A is at (−4, 0). The segment formed by FA
represents one of the legs of the isosceles triangle
formed by F, A, and the point of tangency. Find the
length of this leg.
d=
4(p + 4)(p − 1) = 0
d=
d =5
The solutions to this equation are p = –4 and 1.
Since the parabola opens to the right, p is positive;
thus, p = 1. Use p to find h. h = 1 − (1) or 0. The
vertex of the parabola is (0, 0). Use the vertex and
p to write the standard form of the equation.
4p (x − h) = (y − k)2
4(1)(x − 0) = (y − 0)2
4x = y 2
The lengths of the legs of the isosceles triangle are
5 units. This is the distance from F to the point of
tangency (x, y). Use the distance formula and the
tangent line to solve for the point of tangency.
d
=
5 =
25 = (x − 1)2 + y 2
25 = x2 −2x + 1 + y 2
2
25 = x − 2x + 1 +
2
25 = x − 2x + 1 +
25 =
20 =
2
x + 2x + 4
2
x +5
x
75.
2
SOLUTION:
16 = x2
±4 = x
Find point A, the point of intersection of the axis
and the line tangent to the parabola.
The solutions to this equation are x = ±4. Since A
has an x–coordinate of −4, the point of tangency
must have an x–coordinate of 4. Substitute x = 4 to
solve for y.
The equation for the tangent line is y =
Since point A lies on the x–axis, the y–coordinate is
0. Substitute y = 0 to find x.
y=
x+2
y
=
x+6
y=
(4) + 2
0 =
x+6
y =4
−6 =
The point of tangency is located at (4, 4). Since the
parabola opens to the right, the standard form of
2
the equation is 4p (x − h) = (y − k) and the focus is
located at (h + p , k). The focus is at the point (1,
0); thus, k = 0 and h + p = 1 or h = 1 − p . Use, k,
the standard form of the equation, the point of
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x + 6.
x
−12 = x
A is at (−12, 0). The segment formed by FA Page 36
represents one of the legs of the isosceles triangle
formed by F, A, and the point of tangency. Find the
length of this leg.
0 =
x+6
−6 =
focus is located at (h + p , k). The focus is at the
point (3, 0); thus, k = 0 and h + p = 3 or h = 3 − p .
Use, k, the standard form of the equation, the point
of tangency, and the equation for h to solve for p .
4p (x − h) = (y − k)2
x
7-1 Parabolas
−12 = x
A is at (−12, 0). The segment formed by FA
represents one of the legs of the isosceles triangle
formed by F, A, and the point of tangency. Find the
length of this leg.
4p (12 − h) = (12 − 0)2
48p − 4ph = 144
48p − 4p (3 − p ) = 144
2
48p − 12p + 4p = 144
d=
2
4p + 36p − 144 = 0
4(p + 12)(p − 3) = 0
d=
d = 15
The solutions to this equation are p = −12 and 3.
Since the parabola opens to the right, p is positive;
thus, p = 3. Use p to find h. h = 3 − (3) or 0. The
vertex of the parabola is (0, 0). Use the vertex and
p to write the standard form of the equation.
4p (x − h) = (y − k)2
The lengths of the legs of the isosceles triangle are
15 units. This is the distance from F to the point of
tangency (x, y). Use the distance formula and the
tangent line to solve for the point of tangency.
d
=
4(3)(x − 0) = (y − 0)2
12x = y 2
15 =
225 = (x − 3)2 + y 2
225 = x2 −6x + 9 + y 2
225 = x2− 6x + 9 +
225 = x2 − 6x + 9 +
2
x + 6x + 36
2
225 =
x + 45
180 =
x
76.
2
SOLUTION:
144 = x2
±12 = x
The solutions to this equation are x = ±12. Since A
has an x–coordinate of −12, the point of tangency
must have an x–coordinate of 12. Substitute x = 12
to solve for y.
y=
x+6
y=
(12) + 6
Find point A, the point of intersection of the axis
and the line tangent to the parabola.
The equation for the tangent line is y = 4x + 3.
Since point A lies on the y–axis, the x–coordinate is
0. Substitute x = 0 to find y.
y = 4x + 3
y = 4(0) + 3
y =3
A is at (0, 3). The segment formed by FA
represents one of the legs of the isosceles triangle
formed by F, A, and the point of tangency. Find the
length of this leg.
y = 12
The point of tangency is located at (12, 12). Since
the parabola opens to the right, the standard form
2
of the equation is 4p (x − h) = (y − k) and the
focus is located at (h + p , k). The focus is at the
point (3, 0); thus, k = 0 and h + p = 3 or h = 3 − p .
Use, k, the standard form of the equation, the point
of tangency, and the equation for h to solve for p .
4p (x − h) = (y − k)2
4p (12 − h) = (12 − 0)2
48p − 4ph = 144
48p − 4p (3 − p ) = 144
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2
d
d
d
=
=
=
or 2.125
The lengths of the legs of the isosceles triangle are
Page 37
or 2.125 units. This is the distance from F to
the point of tangency (x, y). Use the distance
=
d
(x − h)
or 2.125
=
2
2
= 4p (y − k)
= 4p (−1 − k)
(−1 − 0)
1 = −4p − 4p k
7-1 Parabolas
1 =
The lengths of the legs of the isosceles triangle are
or 2.125 units. This is the distance from F to
1 = −4p −
the point of tangency (x, y). Use the distance
formula and the tangent line to solve for the point of
tangency.
0 = 4p −
d
2
2
p −1
0 = 8p 2 − 15p − 2
0 = (8p + 1)(p − 2)
=
=
+ 4p
The solutions to this equation are p = 2 and
2
=x +
.
Since the parabola opens down, p is negative; thus,
2
=x +y
2
2
= x + (4x + 3)
2
p=
+
2
vertex of the parabola is (0, 1). Use the vertex and
p to write the standard form of the equation.
2
(x − h) = 4p (y − k)
+
2
= x + 16x + 24x + 9 − 7x −
or 1. The
. Use p to find k.
+
2
=
2
=
(x − 0)
x
2
= 17x + 17x +
0 = 17x2 + 17x
0 = 17x(x + 1)
77. MULTIPLE REPRESENTATIONS In this
The solutions to this equation are x = 0 and −1.
Since A has an x–coordinate of 0, the point of
tangency must have an x–coordinate of −1.
Substitute x = −1 to solve for y.
y = 4x + 3
y = 4(−1) + 3
y = −1
The point of tangency is located at (−1, −1). Since
the parabola opens down, the standard form of the
2
equation is (x − h) = 4p (y − k) and the focus is
located at (h, k + p ). The focus is at the point
; thus, h = 0 and k + p =
or k =
−p .
Use, h, the standard form of the equation, the point
of tangency, and the equation for k to solve for p .
(x − h)
2
2
= 4p (y − k)
= 4p (−1 − k)
(−1 − 0)
1 = −4p − 4p k
1 = −4p −
2
0 = 4p −
2
parabola that has the same vertex as (x + 1) = 20
(y + 7), but is narrower.
e. ANALYTICAL Make a conjecture about the
2
2
2
graphs of x = −2(y + 1), x = −12(y + 1), and x =
−5(y + 1). Check your conjecture by graphing the
parabolas.
SOLUTION:
1 =
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problem, you will investigate how the shape of a
parabola changes as the position of the focus
changes.
a. GEOMETRIC Find the distance between the
vertex and the focus of each parabola.
i. y 2 = 4(x – 2)
2
ii. y = 8(x – 2)
iii. y 2 = 16(x – 2)
b. GRAPHICAL Graph the parabolas in part a
using a different color for each. Label each focus.
c. VERBAL Describe the relationship between a
parabola's shape and the distance between its
vertex and focus.
d. ANALYTICAL Write an equation for a
+ 4p
p −1
2
a.
2
i. Written in standard form, (y − k) = 4p (x − h), p
is the distance between the vertex and the focus.
2
2
y = 4(x − 2) can be written as y = 4(1)(x − 2),
Page 38
where p = 1. Thus, the distance between the vertex
and the focus is 1 unit.
2
parabolas.
(x + 1)
SOLUTION:
7-1
a.
Parabolas
2
i. Written in standard form, (y − k) = 4p (x − h), p
is the distance between the vertex and the focus.
2
2
y = 4(x − 2) can be written as y = 4(1)(x − 2),
where p = 1. Thus, the distance between the vertex
and the focus is 1 unit.
2
ii. Written in standard form, (y − k) = 4p (x − h), p
is the distance between the vertex and the focus.
2
2
y = 8(x − 2) can be written as y = 4(2)(x − 2),
where p = 2. Thus, the distance between the vertex
and the focus is 2 units.
2
iii. Written in standard form, (y − k) = 4p (x − h),
p is the distance between the vertex and the focus.
2
2
y = 16(x − 2) can be written as y = 4(4)(x − 2),
where p = 4. Thus, the distance between the vertex
and the focus is 4 units.
e . Sample answer: The parabolas all have a vertex
of (0, −1) and open downward. The first equation
produces the most narrow parabola and the second
equation produces the widest parabola.
78. ERROR ANALYSIS Abigail and Jaden are
2
graphing x + 6x – 4y + 9 = 0. Is either of them
correct? Explain your reasoning.
b. Use a table of values to graph each parabola.
SOLUTION:
Write the equation in standard form. Since x is
2
squared, it will be in the form (x − h) = 4p (y − k).
x2 + 6x − 4y + = 0
9
2
x + 6x + 9 = 4y
2
(x + 3) = 4y
c. As the focus is moved farther away from the
vertex, the parabola becomes wider.
2
d. The parabola (x + 1) = 20(y + 7) can be written
2
as (x + 1) = (4)(5)(y + 7), where p = 5 and the
vertex is (−1, −7). A parabola that is more narrow
has a smaller value for p . Let p = 1 and write the
standard form for the equation of a parabola with
the same vertex.
2
(x − h) = 4p (y − k)
Abigail; since p = 1, the parabola should open
upward rather than downward.
79. CHALLENGE The area of the parabolic sector
shaded in the graph below is given by
A=
xy. Find the equation of the parabola if the
sector area is 2.4 square units and the width of the
sector is 3 units.
2
= 4(1)[y − (−7)]
2
(x + 1) = 4(y + 7)
[x − (−1)]
e . Sample answer: The parabolas all have a vertex
of (0, −1) and open downward. The first equation
produces the most narrow parabola and the second
equation produces the widest parabola.
SOLUTION:
The width of the sector is 3 units, so y = 1.5. A =
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78. ERROR ANALYSIS Abigail and Jaden are
xy and the area of the sector is 2.4 square units.
Page 39
Substitute the values into the equation and solve for
x.
Sample answer: A point on a parabola is equidistant
from the focus and the directrix. Since a vertex is
directly between the focus and the directrix along
the axis of symmetry, it is the point on a parabola
that is closest to both.
7-1 Parabolas
SOLUTION:
The width of the sector is 3 units, so y = 1.5. A =
xy and the area of the sector is 2.4 square units.
Substitute the values into the equation and solve for
x.
2.4 =
x(1.5)
2.4 = 2x
1.2 = x
Therefore, the vertex of the parabola is (0, 0), and
two other points on the parabola are (1.2, 1.5) and
(1.2, −1.5).
Set up and solve a system of equations using y as
the independent variable.
2
ay + by + c = x
2
a(0) + b(0) + c = 0
2
a(1.5) + b(1.5) + c
2
a(−1.5) + b(−1.5) + c
c
2.25a + 1.5b
2.25a − 1.5b
4.5a
a
2.25 ·
= 1.2
= 1.2
=0
= 1.2
= 1.2
= 2.4
= 1.2
1.2 + 1.5b
1.5b
b
= 1.2
=0
=0
The equation of the parabola is
2
which quadrants the graph of (y – 5) = −8(x + 2)
will have no points. Explain your reasoning.
SOLUTION:
Quadrants I and IV; the vertex is (−2, 5) and p =
−2. Since the vertex is to the left of the y–axis, and
the parabola opens to the left, no points will be to
the right of the y–axis, or in Quadrants I and IV.
82. Writing in Math The concavity of a function’s
graph describes whether the graph opens upward
(concave up) or downward (concave down).
Explain how you can determine the concavity of a
parabola given its focus and vertex.
SOLUTION:
If the focus and the vertex have the same x–
coordinate, then they are concave up or concave
down. If the y–coordinate of the vertex is less than
the y–coordinate of the focus, then the parabola is
concave up. If it is greater than the y–coordinate of
the focus, then the parabola is concave down.
=
+ 1.5b
81. REASONING Without graphing, determine in
If the focus and the vertex have the same y–
coordinate, then they are concave right or concave
left. If the x–coordinate of the vertex is less than
the x–coordinate of the focus, then the parabola is
concave right. If it is greater than the x–coordinate
of the focus, then the parabola is concave left.
2
2
y = x or y =
83. PREWRITE Write a letter outlining and
explaining the content you have learned from this
lesson. Make an outline that addresses purpose,
audience, a controlling idea, logical sequence, and
time frame for completion.
x.
80. REASONING Which point on a parabola is
closest to the focus? Explain your reasoning.
SOLUTION:
Sample answer: A point on a parabola is equidistant
from the focus and the directrix. Since a vertex is
directly between the focus and the directrix along
the axis of symmetry, it is the point on a parabola
that is closest to both.
81. REASONING Without graphing, determine in
2
which quadrants the graph of (y – 5) = −8(x + 2)
will have no points. Explain your reasoning.
SOLUTION:
Quadrants I and IV; the vertex is (−2, 5) and p =
−2. Since the vertex is to the left of the y–axis, and
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the parabola opens to the left, no points will be to
the right of the y–axis, or in Quadrants I and IV.
SOLUTION:
Sample answer:
Purpose: To outline and explain the process of
graphing parabolas.
Audience: Parents of PreCalculus students.
Controlling idea: Using characteristics of
parabolas to graph them.
Logical sequence:
Find the vertex of the parabola.
Find the focus of the parabola.
Find the directrix of the parabola.
Find the axis of symmetry of the parabola.
Graph those characteristics.
Make a table of values and graph the general shape
of the curve.
Timeframe: You have one day to complete the
Page 40
outline.
Find the maximum and minimum values of the
7-1
coordinate, then they are concave right or concave
left. If the x–coordinate of the vertex is less than
the x–coordinate of the focus, then the parabola is
concave right. If it is greater than the x–coordinate
Parabolas
of the focus, then the parabola is concave left.
83. PREWRITE Write a letter outlining and
explaining the content you have learned from this
lesson. Make an outline that addresses purpose,
audience, a controlling idea, logical sequence, and
time frame for completion.
SOLUTION:
Sample answer:
Purpose: To outline and explain the process of
graphing parabolas.
Audience: Parents of PreCalculus students.
Controlling idea: Using characteristics of
parabolas to graph them.
Logical sequence:
Find the vertex of the parabola.
Find the focus of the parabola.
Find the directrix of the parabola.
Find the axis of symmetry of the parabola.
Graph those characteristics.
Make a table of values and graph the general shape
of the curve.
Timeframe: You have one day to complete the
outline.
Find the maximum and minimum values of the
objective function f (x, y) and for what values of
x and y they occur, subject to the given
constraints.
84. x ≤ 5
y ≥ −2
y ≤x–1
f (x, y) = x – 2y
SOLUTION:
Begin by graphing the given system of three
inequalities. The solution of the system, which
makes up the set of feasible solutions for the
objective function, is the shaded region, including its
boundary segments.
The polygonal region of feasible solutions has three
vertices. One vertex is located at (5, −2).
Solve the two systems of two boundary equations
below to find the coordinates of the remaining
vertices.
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by Cognero
System
of
y =x−1
y =x−1
Boundary
x =5
y = −2
Equations
outline.
Find the maximum and minimum values of the
objective function f (x, y) and for what values of
x and y they occur, subject to the given
constraints.
84. x ≤ 5
y ≥ −2
y ≤x–1
f (x, y) = x – 2y
SOLUTION:
Begin by graphing the given system of three
inequalities. The solution of the system, which
makes up the set of feasible solutions for the
objective function, is the shaded region, including its
boundary segments.
The polygonal region of feasible solutions has three
vertices. One vertex is located at (5, −2).
Solve the two systems of two boundary equations
below to find the coordinates of the remaining
vertices.
System of
y =x−1
y =x−1
Boundary
x =5
y = −2
Equations
To solve the first system, substitute x = 5 into the
boundary equation and solve for y.
y=x−1
y = (5) − 1
y =4
A vertex is located at (5, 4).
To solve the second system, substitute y = −2 into
the boundary equation and solve for x.
y =x−1
(−2) = x − 1
−1 = x
A vertex is located at (−1, −2).
Find the value of the objective function f (x, y) = x −
2y at each of the three vertices.
f (5, −2) = (5) − 2(−2) or 9
f (5, 4) = (5) − 2(4) or −3
f (−1, −2) = (−1) − 2(−2) or 3
So, the maximum value of f is 9 when x = 5 and y =
−2. The minimum value of f is −3 when x = 5 and y
= 4.
85. y ≥ −x + 2
2≤x≤7
y≤
x+5
f (x, y) = 8x + 3y
Page 41
So, the maximum value of f is 9 when x = 5 and y =
−2. The minimum value of f is −3 when x = 5 and y
= 4.
7-1 Parabolas
85. y ≥ −x + 2
2≤x≤7
y≤
x+5
A vertex is located at (2, 6).
To solve the second system, substitute x = 7 into
the boundary equation and solve for y.
y=
x+5
y=
(7) + 5
y = 8.5
A vertex is located at (7, 8.5).
f (x, y) = 8x + 3y
SOLUTION:
Begin by graphing the given system of three
inequalities. The solution of the system, which
makes up the set of feasible solutions for the
objective function, is the shaded region, including its
boundary segments.
To solve the third system, substitute x = 2 into the
boundary equation and solve for y.
y = −x + 2
y = −(2) + 2
y =0
A vertex is located at (2, 0).
To solve the fourth system, substitute x = 7 into the
boundary equation and solve for y.
y = −x + 2
y = −(7) + 2
y = −5
A vertex is located at (7, −5).
The polygonal region of feasible solutions has four
vertices.
Solve the four systems of two boundary equations
below to find the coordinates of the remaining
vertices.
System of
Boundary
Equations
System of
Boundary
Equations
y=
x+
y=
x+
5
x =2
5
x =7
y = −x + 2
x =2
y = −x + 2
x =7
Find the value of the objective function f (x, y) = 8x
+ 3y at each of the four vertices.
f (2, 6) = 8(2) + 3(6) or 34
f (7, 8.5) = 8(7) + 3(8.5) or 81.5
f (2, 0) = 8(2) + 3(0) or 16
f (7, −5) = 8(7) + 3(−5) or 41
So, the maximum value of f is 81.5 when x = 7 and
y = 8.5. The minimum value of f is 16 when x = 2
and y = 0.
86. x ≥ −3
y ≥1
3x + y ≤ 6
f (x, y) = 5x – 2y
SOLUTION:
To solve the first system, substitute x = 2 into the
boundary equation and solve for y.
y=
x+5
y=
(2) + 5
Begin by graphing the given system of three
inequalities. The solution of the system, which
makes up the set of feasible solutions for the
objective function, is the shaded region, including its
boundary segments.
y =6
A vertex is located at (2, 6).
To solve the second system, substitute x = 7 into
the boundary equation and solve for y.
y=
x+5
y=
(7) + 5
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y = 8.5
A vertex is located at (7, 8.5).
The polygonal region of feasible solutions has three
Page 42
vertices. One vertex is located at (−3, 1)
Solve the two systems of two boundary equations
below to find the coordinates of the remaining
f (1.67, 1) = 5(1.67) − 2(1) or 6.35
So, the maximum value of f is 6.35 when x = 1.67
and y = 1. The minimum value of f is −45 when x =
−3 and y = 15.
7-1 Parabolas
The polygonal region of feasible solutions has three
vertices. One vertex is located at (−3, 1)
Solve the two systems of two boundary equations
below to find the coordinates of the remaining
vertices.
System of
3x + y = 6 3x + y = 6
Boundary
x = −3
y =1
Equations
To solve the first system, substitute x = −3 into the
boundary equation and solve for y.
3x + y = 6
3(−3) + y = 6
−9 + y = 6
y = 15
A vertex is located at (−3, 15).
To solve the second system, substitute y = 1 into
the boundary equation and solve for x.
A vertex is located at (1.67, 1).
3x + y = 6
3x + (1) = 6
3x = 5
x ≈ 1.67
Find the value of the objective function f (x, y) = 5x
− 2y at each of the three vertices.
f (−3, 1) = 5(−3) − 2(1) or −17
f (−3, 15) = 5(−3) − 2(15) or −45
f (1.67, 1) = 5(1.67) − 2(1) or 6.35
So, the maximum value of f is 6.35 when x = 1.67
and y = 1. The minimum value of f is −45 when x =
−3 and y = 15.
87. Find the partial fraction decomposition of
.
SOLUTION:
87. Find the partial fraction decomposition of
.
SOLUTION:
Rewrite the equation as partial fractions with
constant numerators, A and B, and denominators
that are the linear factors of the original
denominator.
=
+
2y + 5 = A(y + 1) + B(y + 2)
2y + 5 = Ay + A + By + 2B
2y + 5 = (A + B)y + (A + 2B)
Equate the coefficients on the left and right side of
the equation to obtain a system of two equations.
A +B =2
A + 2B = 5
To solve the system, let A = 2 − B and use
substitution.
(2 − B) + 2B = 5
2+B =5
B =3
Thus, A = 2 − (3) or −1. So, the partial fraction
decomposition of
.
88. SURVEYING Talia is surveying a rectangular lot
for a new office building. She measures the angle
between one side of the lot and the line from her
position to the opposite corner of the lot as
.
She then measures the angle between that line and
the line to a telephone pole on the edge of the lot as
. If Talia is 100 yards from the opposite corner
of the lot, how far is she from the telephone pole?
Rewrite the equation as partial fractions with
constant numerators, A and B, and denominators
that are the linear factors of the original
denominator.
=
+
2y + 5 = A(y + 1) + B(y + 2)
2y + 5 = Ay + A + By + 2B
2y + 5 = (A + B)y + (A + 2B)
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Equate the coefficients on the left and right side of
the equation to obtain a system of two equations.
A +B =2
SOLUTION:
Make a sketch of the drawing.
Page 43
= 50
y cos
y
=
y = 51.76
Talia is about 51.76 yards from the telephone pole.
7-1 Parabolas
SOLUTION:
Make a sketch of the drawing.
Find the value of each expression using the
given information.
89. cot θ and csc θ; tan θ =
, sec θ > 0
SOLUTION:
Use the reciprocal identity to find cot θ.
cot θ =
cot θ =
The rectangle is divided into three triangles with the
last triangle having a known angle of 15° found by
subtracting
from
. Separating the
and
triangles, we have a right triangle with a side length
of 100 yards and an angle of
.
cot θ =
Use the Pythagorean identity and cot θ to find csc
θ.
2
2
cot θ + 1 = csc θ
+ 1 = csc2 θ
2
+
= csc θ
2
= csc θ
Solve for x using trigonometric properties.
cos
= csc θ
±
=
= csc θ
±
=x
50 = x
Use the value of x as a side length of the right
triangle that has y as a side and
as an angle.
100 cos
Since sec θ > 0 and cot is positive, csc > 0. Thus,
csc =
.
90. cos θ and tan θ; csc θ = −2, cos θ < 0
SOLUTION:
Solve for y using trigonometric properties.
=
cos
sin θ =
= 50
y cos
y
=
y = 51.76
Talia is about 51.76 yards from the telephone pole.
Find the value of each expression using the
given information.
89. cot θ and csc θ; tan θ =
Use the reciprocal identity to first find sin θ.
sin θ =
, sec θ > 0
SOLUTION:
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Use the Pythagorean identity and sin θ to find cos
θ.
2
2
sin θ + cos θ = 1
2
+ cos θ = 1
2
=1
2
=
+ cos θ
cos θ
Page 44
Use the reciprocal identity to find cot θ.
cos θ
=±
or ±
±
tan θ =
Since sec θ > 0 and cot is positive, csc > 0. Thus,
csc =
.
7-1 Parabolas
tan θ =
or
Locate the vertical asymptotes, and sketch
the graph of each function.
91. y = tan x + 4
90. cos θ and tan θ; csc θ = −2, cos θ < 0
SOLUTION:
Use the reciprocal identity to first find sin θ.
sin θ =
SOLUTION:
The graph of y = tan θ + 4 is the graph of y = tan x
sin θ =
Use the Pythagorean identity and sin θ to find cos
θ.
2
2
sin θ + cos θ = 1
location of two consecutive vertical asymptotes.
2
+ cos θ = 1
2
=1
2
=
+ cos θ
cos θ
cos θ
=±
Since cos θ < 0, cos θ = −
and
Create a table listing the coordinates of key points
for y = tan θ + 4 for one period on
or ±
.
Function
Use the quotient identity, cos θ, and sin θ to find tan
θ.
tan θ =
tan θ =
tan θ =
or π. Find the
shifted 4 units up. The period is
Locate the vertical asymptotes, and sketch
the graph of each function.
91. y = tan x + 4
SOLUTION:
y = tan x
y = tan x +
4
(0, 0)
(0, 4)
Vertical
Asymptote
Intermediate
Point
x-intercept
or
.
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points
for the function. Then repeat the pattern to sketch
more to the left and right of the first curve.
The graph of y = tan θ + 4 is the graph of y = tan x
shifted 4 units up. The period is
or π. Find the
location of two consecutive vertical asymptotes.
and
92. y = sec x + 2
SOLUTION:
The graph of y = sec x + 2 is the graph of y = sec x
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Create a table listing the coordinates of key points
shifted 2 units up. The period is
for y = tan θ + 4 for one period on
location of two vertical asymptotes.
.
Page 45
or 2π. Find the
7-1 Parabolas
92. y = sec x + 2
93. y = csc x −
SOLUTION:
The graph of y = sec x + 2 is the graph of y = sec x
or 2π. Find the
shifted 2 units up. The period is
location of two vertical asymptotes.
SOLUTION:
The graph of y = csc x −
x shifted
is the graph of y = csc
unit down. The period is
or π. Find
the location of two consecutive vertical asymptotes.
and
and
Create a table listing the coordinates of key points
for y = sec x + 2 for one period on
.
Create a table listing the coordinates of key points
Function
Vertical
Asymptote
Intermediate
Point
y = sec x
y = sec x +
2
for y = csc x –
for one period on
.
y = csc x –
(0, 1)
(0, 3)
Vertical
Asymptote
Function
Vertical
Asymptote
Intermediate
Point
Intermediate
Point
Vertical
Asymptote
Vertical
Asymptote
Sketch the curve through the indicated key points
for the function. Then repeat the pattern to sketch
at least one more cycle to the left and right of the
first curve.
Intermediate
Point
Vertical
Asymptote
y = csc x
x = −π
x = −π
x =0
x=π
x=π
Sketch the curve through the indicated key points
for the function. Then repeat the pattern to sketch
more to the left and right of the first curve.
93. y = csc x −
SOLUTION:
The graph of y = csc x −
is the graph of y = csc
x shifted
unit down. The period is
Evaluate each expression.
94. log16 4
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or π. Find
SOLUTION:
Page 46
SOLUTION:
x
log3 27
3 x
= log3 (3 )
3x
= log3 3
= 3x
7-1 Parabolas
Evaluate each expression.
Graph each function.
97. f (x) = x3 – x2 – 4x + 4
94. log16 4
SOLUTION:
SOLUTION:
Use a table of values to graph the function.
log16 4 = y
y
x
−3
−2
−1
0
1
2
3
=4
16
2y
1
(4 ) = 4
2y
1
4
=4
Solve 2y = 1 for y.
2y = 1
y
=
f(x)
−20
0
6
4
0
0
10
95. log4 16x
SOLUTION:
x
2 x
log4 16
= log4 (4 )
2x
= log4 4
= 2x
96. log3 27x
SOLUTION:
x
3 x
log3 27
= log3 (3 )
3x
= log3 3
= 3x
Graph each function.
3
2
97. f (x) = x – x – 4x + 4
SOLUTION:
Use a table of values to graph the function.
x
−3
−2
−1
0
1
2
3
f(x)
−20
0
6
4
0
0
10
98. g(x) = x4 – 7x2 + x + 5
SOLUTION:
Use a table of values to graph the function.
x
−3
−2
−1
0
1
2
3
f(x)
20
−9
−1
5
0
−5
26
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Page 47
4
2
7-1 Parabolas
98. g(x) = x4 – 7x2 + x + 5
100. REVIEW What is the solution set for 3(4x + 1)2
= 48?
SOLUTION:
A
Use a table of values to graph the function.
x
−3
−2
−1
0
1
2
3
f(x)
20
−9
−1
5
0
−5
26
B
C
D
E
SOLUTION:
2
3(4x + 1) = 48
2
(4x + 1) = 16
4x + 1 = ± 4
Solve the left side of the equation for both 4 and
−4.
4x + 1 = 4
4x = 3
x
4x + 1 = −4
4x = −5
99. h(x) = x4 – 4x2 + 2
SOLUTION:
x
Use a table of values to graph the function.
x
−3
−2
−1
0
1
2
3
=
f(x)
47
2
−1
2
−1
2
47
=
The solution set for the equation is x =
and
.
101. SAT/ACT If x is a positive number, then
F
G
H
J
SOLUTION:
=
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100. REVIEW
WhatbyisCognero
the solution
= 48?
A
2
=
set for 3(4x + 1)
Page 48
=
x
= −5
Dy =x
=
SOLUTION:
7-1 Parabolas
The solution set for the equation is x =
and
101. SAT/ACT If x is a positive number, then
.
The graph is a parabola. The parent function for a
2
parabola is y = x .
103. REVIEW What are the x–intercepts of the
2
graph of y = −2x – 5x + 12?
F
F
G −4,
G
H −2,
H
J
J
SOLUTION:
SOLUTION:
=
The x–intercepts of a graph occur at the zeros of
2
the function. Factor to solve 0 = −2x − 5x + 12 for
x.
2
=
=
102. Which is the parent function of the graph shown
below?
0 = −2x − 5x + 12
0 = (−2x + 3)(x + 4)
The solutions to this equation occur when either of
the factors is equal to 0.
−2x + 3 = 0
−2x = −3
x
=
x +4 =0
x = −4
The x–intercepts of the graph of y are −4 and
Ay
By
Cy
Dy
.
=x
=
= |x|
=x
2
SOLUTION:
The graph is a parabola. The parent function for a
2
parabola is y = x .
103. REVIEW What are the x–intercepts of the
2
graph of y = −2x – 5x + 12?
F
G −4,
H −2,
J
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SOLUTION:
The x–intercepts of a graph occur at the zeros of
Page 49