MATH 150
Quiz Key #6
10/27-29/2015
(1) Find the domain, x-intercepts, and y-intercepts for f ( x ) =
5x3 −13x2 −6x
.
x3 −3x2 −4x +12
Solution: Start by factoring the numerator and denominator:
x 5x2 − 13x − 6
x (5x + 2) ( x − 3)
x (5x + 2) ( x − 3)
f (x) = 2
=
=
.
2
( x + 2) ( x − 2) ( x − 3)
x ( x − 3) − 4 ( x − 3)
( x − 4) ( x − 3)
Now, before cancelling anything out, we find the domain. In this case, we just
have to check that the denominator is not zero. Hence, x 6= ±2, 3. Therefore,
the DOMAIN is given by (−∞, −2) ∪ (−2, 2) ∪ (2, 3) ∪ (3, ∞) . Now, provided
you are in the domain,
x (5x + 2)
f (x) =
.
( x + 2) ( x − 2)
To find the x-intercept, we solve the following equation:
0=
x (5x + 2)
( x + 2) ( x − 2)
0 = x (5x + 2)
2
x = 0 or x = − .
5
Therefore, the x-INTERCEPTS are (0, 0) and − 25 , 0 . To find the y-intercept,
we substitute 0 for x. Therefore, the y-INTERCEPT is (0, 0) .
(2) Using the generalized technique, determine the end behavior of f ( x ) =
x7 −5x10 +7x −9
.
6x3 +13x12 +9x2
Solution: The highest power of x is x12 . Therefore, we do the following:
1
1
7
10
− x52 + x711 − x912
x − 5x + 7x − 9
x12
x5
f (x) =
·
=
.
6
1
6x3 + 13x12 + 9x2
+ 13 + x910
x9
12
x
Hence, as x → ±∞,
f (x) →
0−0+0−0
0
=
= 0.
0 + 13 + 0
13
1
(3) Find the horizontal asymptotes, if there are any, for the following equations. [Use the theorem.]
(a)
7x3 + 3x94 − 5x56
f (x) =
16x4 − 14x97 + 7x74
(b)
f (x) =
x10 + 8 − 15x27
2x14 + 7x27 − 4
(c)
f (x) =
7x − 5x4
3x2 − 6x + 9
(a) Since the highest power is 97, the horizontal asymptote is y = 0.
(b) Since the highest power is 27, the horizontal asymptote is y = − 15
7 .
(c) Since the highest power is 4, there is no horizontal asymptote.
(4) Use the graph of f ( x ) provided to answer all the questions.
Domain: (−∞, −5) ∪ (−5, −4) ∪ (−4, 0) ∪ (0, ∞)
Range: (−∞, ∞)
x-intercepts: (−3, 0) ; (4, 0)
y-intercepts: none
Horizontal Asymptote(s): y = 1
Vertical Asymptote(s): x = −5, x = 0
Hole(s): (−4, −2)
2
(5) Find all vertical asymptotes and holes for the function f ( x )
2x3 −3x2 −35x
. Also, find any zeros of the function.
3x4 −19x3 +20x2
=
First, we factor:
f (x) =
x 2x2 − 3x − 35
x (2x + 7) ( x − 5)
= 2
.
x2 (3x2 + 19x + 20)
x (3x − 4) ( x − 5)
When x 6= 5, this is
f (x) =
2x + 7
.
x (3x − 4)
4
Therefore, the VERTICAL ASYMPTOTES of f ( x ) are x =
0 and
x = 3 . The
17
HOLE is found at x = 5, so plugging in, we get the point 5, 55
. Finally, the
ZERO is the solution to the equation:
0=
2x + 7
x (3x − 4)
7
0 = 2x + 7 =⇒ x = − .
2
Therefore, the ZERO is − 72 .
(6) For the following angles, determine the reference angle in radians and
whether x and y will be positive, negative, or zero, where x and y are the
terminating side of the angle that intersects a circle centered at the origin.
Angle
Ref. Angle
π
3
3π
4
7π
6
5π
3
3π
2
13π
4
π
3
π
4
π
6
π
3
π
2
π
4
x
pos
neg
neg
pos
0
neg
y
pos
pos
neg
neg
neg
neg
3