MATH 150
Quiz Key #4
10/6-8/2015
(1) Graph the equation − x2 + 2x + y + 3 = 0 by plotting points. Plot at least
5 points and your graph should include all x- and y-intercepts. Your graph
must also have the correct shape.
Solution: To find the x-intercepts, solve the following equation:
− x2 + 2x + 3 = 0 =⇒ x2 − 2x − 3 = 0 =⇒ ( x − 3) ( x + 1) = 0.
Therefore, the x-intercepts are at (−1, 0) and (3, 0) . To find the y-intercepts,
solve the equation: y + 3 = 0. Therefore, the y-intercept is (0, −3) . We need
two more points, so plug in x = 1 and x = 2 to get the y-values:
−1 + 2 + y + 3 = 0 =⇒ y = −4 and − 4 + 4 + y + 3 = 0 =⇒ y = −3.
Therefore, our five points are (−1, 0) ; (0, −3) ; (1, −4) ; (2, −3) ; and (3, 0) . Now
sketch the parabola through those points.
(2) Find the x- and y-intercepts for the equation: x2 y3 + 10y + 2 | x − 3| + yx2 −
3y2 = 14 + x4 y4 .
Solution: To find the x-intercepts, we plug in 0 for y to obtain the equation:
x2 · 0 + 10 · 0 + 2 | x − 3| + 0x2 − 3 · 0 = 14 + x4 · 0.
2 | x − 3| = 14 =⇒ | x − 3| = 7 =⇒ x − 3 = ±7.
Thus, x = 3 ± 7. Hence, x = 10 or x = −4. Therefore, the x-intercepts are
(10, 0) and (−4, 0) . Now to find the y-intercepts, we plug in 0 for x to obtain
the equation:
0 · y3 + 10y + 2 |0 − 3| + y · 0 − 3y2 = 14 + 0 · y4 .
10y + 6 − 3y2 = 14 =⇒ 0 = 3y2 − 10y + 8 =⇒ 0 = (3y − 4) (y − 2) .
Therefore, y = 43 or y = 2. Therefore, the y-intercepts are 0, 43 and (0, 2) .
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(3) Test the equation xy3 = x2 | xy| − xy for symetry about the x-axis, y-axis,
and origin.
Solution: (a) To test symmetry about the x-axis, plug in −y for y. This gives
the equation:
x (−y)3 = x2 | x (−y)| − x (−y) .
− xy3 = x2 |− xy| + xy.
− xy3 = x2 | xy| + xy.
This is not the same equation, so it is not symmetric about the x-axis.
(b) To test symmetry about the y-axis, plug in − x for x. This gives the equation:
− xy3 = (− x )2 |− xy| + xy.
− xy3 = x2 | xy| + xy.
This is not the same equation, so it is not symmetric about the y-axis.
(c) To test symmetry about the origin, plug in − x for x and −y for y. This
gives the equation:
− x (−y)3 = (− x )2 |− x (−y)| + x (−y) .
xy3 = x2 | xy| − xy.
This is the same equation, so it is symmetric about the origin.
(4) If the line through (5, 6) and (13, y2 ) is parallel to the line 5x − 8y = 16,
find y2 .
Solution: First we need to know the slope of our line. To do this place the
equation in slope-intercept form.
−8y = −5x + 16 =⇒ y =
5
x − 2.
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Hence, the slope of the line is 85 . Since our lines is to be parallel, the slope of
our line also needs to be 58 . Therefore,
y2 − 6
5
= .
13 − 5
8
y2 − 6
5
= .
8
8
y2 − 6 = 5.
Therefore, y2 = 11.
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(5) Find the perpendicular bisector of the line segment between A (−2, 3) and
B (5, −6) . Write your answer in slope-intercept form.
Solution: To find the equation of a line, we need to know its slope and a point
on the line. To find the slope, we first find the slope of AB.
m AB =
9
−6 − 3
=− .
5 − (−2)
7
Therefore,
m⊥ =
7
.
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The perpendicular bisector cuts AB in half. Therefore, it intersects AB at the
midpoint. The midpoint is given by
3 3
−2 + 5 3 − 6
,
=
,−
.
2
2
2 2
Now using point-slope form, we get the following equation for the perpendicular bisector:
3
7
3
y− −
=
x−
.
2
9
2
We now simplify and isolate y to get an equation in slope-intercept form:
y+
3
7
7
= x− .
2
9
6
7 3
7
x− − .
9
6 2
7
7 9
y = x− − .
9
6 6
7
16
y = x− .
9
6
7
8
y = x− .
9
3
y=
(6) Label the two special right triangles. [Label the angles in radians.]
Solution: The first triangle is a π4 - π4 - π2 triangle. The sides opposite the π4 angles
√
are each 1 unit long, while the hypotenuse is 2 units long. The second triangle
is a π6 - π3 - π2 triangle. The side opposite the π6 angle is 1 unit long, the side
√
opposite the π3 angle is 3 units long, and the hypotenuse is 2 units long.
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(7) If the length of the leg across from the
what is the length of the hypotenuse?
π
3
angle of a
π π π
6-3-2
triangle is 4,
Solution: We can find the length of the hypotenuse by setting up the following
proportion (based on similar triangles):
√
√
3
4
8
8 3
√
=
=⇒ h =
=
.
2
h
3
3
Therefore, the length of the hypotenuse is
(8) If the length of one leg of a
hypotenuse and the other leg?
π π π
4-4-2
√
8 3
3 .
triangle is 5, what is the length of the
Solution: The two legs of this triangle have the same length, so the other leg is
5 units long. Now, setting up a proportion, we get:
√
√
h
2
=
=⇒ h = 5 2.
5
1
√
Hence the hypotenuse is 5 2 units long.
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