Chabot Mathematics
§8.3 Quadratic
Fcn Graphs
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Review § 8.2
MTH 55
 Any QUESTIONS About
• §8.2 → Quadratic Eqn Applications
 Any QUESTIONS About HomeWork
• §8.2 → HW-38
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphs of Quadratic Eqns
 All quadratic functions have graphs
similar to y = x2. Such curves are called
parabolas. They are U-shaped and
symmetric with respect to a vertical line
known as the parabola’s line of
symmetry or axis of symmetry.
 For the graph of f(x) = x2, the y-axis is
the axis of symmetry. The point (0, 0) is
known as the vertex of this parabola.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f(x) = 2x2
 Solution:
Make T-Table and
Connect-Dots
x
y
(-2,8)
0
(0, 0)
1
2
(1, 2)
(2,8)
8
7
6
5
4
(x, y)
0
(-1,2 )
3
2
(1,2)
1
-5 -4 -3 -2 -1
1
2 3 4 5
-1
x
–1
2
(–1, 2)
2
8
(2, 8)
 x = 0 is Axis of Symm
–2
8
(–2, 8)
 (0,0) is Vertex
Chabot College Mathematics
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y
-2
(0, 0)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f(x) = −3x2
y
 Solution:
Make T-Table and
Connect-Dots
x
y
6
5
4
3
2
1
(x, y)
-5 -4 -3 -2 -1
0
0
(0, 0)
1
–3
(1, –3)
–1
–3
(–1, –3)
2
–12 (2, –12)
–2
–12 (–2, –12)
Chabot College Mathematics
5
x
1
-1
-2
2 3 4 5
-3
-4
-5
 Same Axis & Vertex
but opens
DOWNward
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Examples of ax2 Parabolas
f ( x)  4 x
2
f ( x)  x 2
6
5
4
1 2
f ( x)  x
4
f ( x)   x
Chabot College Mathematics
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3
2
1
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2
 The graph of f(x) = ax2 is a parabola with
• x = 0 as its axis of symmetry.
• The Origin, (0,0) as its vertex.
 For Positive a the parabola opens upward
 For Negative a the parabola opens downward
 If |a| is greater than 1; e.g., 4, the parabola is
narrower (tighter) than y = x2.
 If |a| is between 0 and 1 e.g., ¼, the parabola
is wider (broader) than y = x2.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
The Graph of f(x) = a(x – h)2
 We could next consider graphs of
f(x) = ax2 + bx + c, where b and c are
not both 0.
 It turns out to be convenient to first
graph f(x) = a(x – h)2, where h is some
constant; i.e., h is a NUMBER
 This allows us to observe similarities to
the graphs drawn in previous slides.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f(x) = (x−2)2
y
 Solution:
Make T-Table and
Connect-Dots
x
y
(x, y)
0
4
8
7
6
5
4
3
2
(0, 4)
1
1
1
–1
9
(–1, 9)
2
0
(2, 0)
3
1
(3, 1)
4
4
(4, 4)
Chabot College Mathematics
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(1, 1)
-5 -4 -3 -2 -1
x
1
2 3 4 5
-1
vertex
-2
 The Vertex SHIFTED
2-Units to the Right
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = a(x−h)2
 The graph of y = f(x) = a(x – h)2 has the
same shape as the graph of y = ax2.
 If h is positive, the graph of y = ax2 is
shifted h units to the right.
 If h is negative, the graph of y = ax2 is
shifted |h| units to the left.
 The vertex is (h, 0) and the axis of
symmetry is x = h.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graph of f(x) = a(x – h)2 + k
 Given a graph of f(x) = a(x – h)2, what
happens if we add a constant k?
 Suppose we add k = 3. This increases f(x)
by 3, so the curve moves up
• If k is negative, the curve moves down.
 The axis of symmetry for the parabola
remains x = h, but the vertex will be at
(h, k), or equivalently (h, f(h))
• f(h) = a([h] – h)2 + k = 0 + k → f(h) = k
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
1
f ( x)   ( x  3)2  1.
2
y
 Make T-Table and
Connect-Dots
x
0
–1
–2
–3
–4
–5
y
1
-5 -4 -3 -2 -1
-1
-2
-11/2 (0, -11/2)
–3
–1
(–3, –1)
–3/2 (–4, –3/2)
–3
(–5, –3)
x
1
2 3 4 5
-3
-4
(–1, –3)
–3/2 (–2, –3/2)
Chabot College Mathematics
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(x, y)
3
2
-5
-6
-7
vertex
-8
 The Vertex SHIFTED
3-Units Left and
1-Unit Down
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Quadratic Fcn in Standard Form
 The Quadratic Function Written in
STANDARD Form:
f x   a x  h   k , a  0
2
 The graph of f is a parabola with vertex
(h, k). The parabola is symmetric with
respect to the line x = h, called the axis
of the parabola. If a > 0, the parabola
opens up, and if a < 0, the parabola
opens down.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Find Quadratic Fcn
 Find the standard form of the quadratic
function whose graph has vertex (−3, 4)
and passes through the point ( −4, 7).
 SOLUTION: Let y = f(x) be the
quadratic function. Then
2
2
7  a –4  3  4
y  a x  h   k
a3
2
y  a x  3  4
Hence,
2
2
y  a x  3  4
y  3 x  3  4
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = a(x – h)2 + k
1. The graph is a parabola.
•
Identify a, h, and k
2. Determine how the parabola opens.
•
If a > 0 (positive), the parabola opens up.
•
If a < 0 (negative), the parabola opens down.
3. Find the vertex. The vertex is (h, k).
•
If a > 0 (or a < 0), the function f has a
minimum (or a maximum) value k at x = h
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = a(x – h)2 + k
4. Find the x-intercepts.
•
Find the x-intercepts (if any) by setting
f(x) = 0 and solving the equation
a(x – h)2 + k = 0 for x.
– Solve by one of: [Factoring + ZeroProducts],
Quadratic Formula, CompleteSquare
– If the solutions are real numbers, they are the
x-intercepts.
– If the solutions are NOT Real Numbers, the
parabola either lies above the x–axis (when a
> 0) or below the x–axis (when a < 0).
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = a(x – h)2 + k
5. Find the y-intercept
•
Find the y-intercept by replacing x with 0.
Then y = f(0) = ah2 + k is the y-intercept.
6. Sketch the graph
•
Plot the points found in Steps 3-5 and join
them by a parabola.
– If desired, show the axis of symmetry, x = h,
for the parabola by drawing a dashed
vertical line
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f x   2 x  3  8.
2
 SOLUTION
Step 1 a = 2, h = 3, and k = –8
Step 2 a = 2, a > 0, the parabola opens up.
Step 3 (h, k) = (3, –8); the function f has a
minimum value –8 at x = 3.
Step 4 Set f (x) = 0 and solve for x.
2
0  2 x  3  8
x  3  2
8  2 x  3
2
4  x  3
2
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x  5 or x  1
x-intercepts: 1 and 5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f x   2 x  3  8.
2
 SOLUTION cont.
Step 5 Replace x with 0.
f 0   2 0  3  8
2
 2 9   8  10
y-intercept is 10 .
Step 6 axis: x = 3, opens up, vertex: (3, –8),
passes through (1, 0), (5, 0) and (0, 10),
the graph is y = 2x2 shifted three units
right and eight units down.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f x   2 x  3  8.
2
 SOLUTION
cont.
• Sketch Graph
Using the 4 points
– Vertex
– Two x-Intercepts
– One y-Intercept
f x   2 x  3  8.
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Completing the Square
 By completing the square, we can
rewrite any polynomial ax2 + bx + c
in the form a(x – h)2 + k.
 Once that has been done, the
procedures just discussed enable
us to graph any quadratic function.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
2
Example  Graph f ( x)  x  2 x  1.
 SOLUTION
f (x) = x2 – 2x – 1
Adding ZERO
y
6
= (x2 – 2x) – 1
5
4
3
= (x2 – 2x + 1 – 1) – 1
2
= (x2 – 2x + 1) – 1 – 1
= (x –
1)2
1
-5 -4 -3 -2 -1
–2
x
1
-1
-2
 The vertex is at (1, −2)
-3
-4
 The Parabola
Opens UP
-5
Chabot College Mathematics
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2 3 4 5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
2
Example  Graph f ( x)  2 x  6 x  3.
 SOLUTION
y
f (x) = –2x2 + 6x – 3
6
Complete
Square
= –2(x2 – 3x) – 3
=
–2(x2
5
4
3
2
1
– 3x + 9/4 – 9/4) – 3
-5 -4 -3 -2 -1
= –2(x2 – 3x + 9/4) – 3 + 18/4
1
-1
-2
x
2 3 4 5
-3
-4
= –2(x – 3/2)2 + 3/2
-5
 The vertex is at (3/2, 3/2)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
The Vertex of a Parabola
 By the Process of Completing-theSquare we arrive at a FORMULA for
the vertex of a parabola given by
f(x) = ax2 + bx + c:
2

b
b
4
ac

b
 b



 
  2a , f  2a   or   2a , 4a .
 



• The x-coordinate of the vertex is −b/(2a).
• The axis of symmetry is x = −b/(2a).
• The second coordinate of the vertex is most
commonly found by computing f(−b/[2a])
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2 + bx + c
1. The graph is a parabola.
Identify a, b, and c
2. Determine how the parabola opens
•
If a > 0, the parabola opens up.
•
If a < 0, the parabola opens down
3. Find the vertex (h, k). Use the formula
 b
 b 
h, k     , f    .
 2a
2a 
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2 + bx + c
4. Find the x-intercepts
Let y = f(x) = 0. Find x by solving the
equation ax2 + bx + c = 0.
•
If the solutions are real numbers,
they are the x-intercepts.
•
If not, the parabola either lies
– above the x–axis when a > 0
– below the x–axis when a < 0
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2 + bx + c
5. Find the y-intercept. Let x = 0. The
result f(0) = c is the y-intercept.
6. The parabola is symmetric with
respect to its axis, x = −b/(2a)
•
Use this symmetry to find
additional points.
7. Draw a parabola through the points
found in Steps 3-6.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 SOLUTION
Step 1 a = –2, b = 8, and c = –5
Step 2 a = –2, a < 0, the parabola opens down.
Step 3 Find (h, k).
b
8
h

2
2a
2 2 
k  f 2   2 2   8 2   5  3
2
h, k   2, 3
Maximum value of y = 3 at x = 2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 SOLUTION
2
2x
 8x  5  0
Step 4 Let f (x) = 0.
x
8 
8 
2
 4 2 5 
2 2 
4 6

2
4 6
4 6
x-intercepts are
and
.
2
2
Step 5
Let x = 0. f 0   2 0   8 0   5
y-intercept is  5 .
Chabot College Mathematics
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2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 SOLUTION
Step 6 Axis of symmetry is x = 2. Let x = 1,
then the point (1, 1) is on the graph, the
symmetric image of (1, 1) with respect to
the axis x = 2 is (3, 1). The symmetric
image of the y–intercept (0, –5) with
respect to the axis x = 2 is (4, –5).
Step 7 The parabola passing through the points
found in Steps 3–6 is sketched on the
next slide.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 SOLUTION cont.
• Sketch Graph
Using the points
Just Determined
f x   2x  8x  5
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Find Domain & Range
 Given the graph of
f(x) = −2x2 +8x − 5
 Find the domain and
range for f(x)
 SOLUTION 
Examine the Graph
to find that the:
• Domain is (−∞, ∞)
• Range is (−∞, 3]
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
WhiteBoard Work
 Problems From §8.3 Exercise Set
• 4, 16, 22, 30
 The Directrix
of a Parabola
• A line perpendicular to the axis
of symmetry used in the definition
of a parabola. A parabola is
defined as follows: For a given point,
called the focus, and a given line
not through the focus, called the
directrix, a parabola is the locus
of points such that the distance to
the focus equals the distance to the directrix.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
All Done for Today
Geometric
Complete
The
Square
x  10 x 
2
x  10 x  25  x  5
2
Chabot College Mathematics
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2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
8
10