Download 3 To find the vertical asymptotes you want to find out where the

#14/p. 382 f ( x ) =
3
x2
To find the vertical asymptotes you want to find out where the denominator is zero.
(This is also what you do to find the domain.)
So x 2 = 0 ⇒ x = 0 which is your vertical asymptote.
To find horizontal or oblique asymptotes you need to see what happens to your function
as x gets very large (ie as x gets close to ∞)
There are some short cuts that you can use on the test and the homework:
IF f ( x ) =
p( x )
then we have three cases:
q( x)
Case 1: degree p(x) < degree q(x) Then the horizontal asymptote is y = 0.
Ex: f ( x ) =
x2 + 7
has horizontal asymptote y = 0.
x3
Case 2: degree p(x) = degree q(x) Then the horizontal asymptote is the leading
coefficients of the functions:
Ex: f ( x ) =
ax 3 + 7
a
the horizontal asymptote is y =
3
2
bx + 54 x − 1009
b
Case 3: degree p(x) > degree q(x) Then there is no horizontal asymptote but there is an
oblique asymptote. To find the oblique asymptote you want to divide the numerator of
the fraction by the denominator. When you are done with the division you will get
r ( x)
where Q( x ) is the quotient and r(x) is
something that looks like: f ( x ) = Q( x ) +
q( x)
the remainder. The oblique asymptote is y = Q( x ) .
For this problem we are in Case 1 so the horizontal asymptote is y = 0.
#18/p.382 Graph f ( x ) =
3x
x − 5x + 6
2
First find your asymptotes: For the horizontal asymptote we are in Case 1 again so y = 0
is the horizontal asymptote. (that is the x-axis)
Second find your vertical asymptote: Set the denominator = 0.
x 2 − 5x + 6 = ( x − 3)( x − 2) = 0 so x = 3 or x = 2 are the vertical asymptotes.
You want to sketch them on your graph. The vertical asymptotes are vertical lines
through x = 3 and x = 2.
Next you want to see what happens to the function as x gets close to your asymptotes. To
do that you plug in numbers into f(x) that are close to 3 and 2. You need to try numbers
on both sides. The only thing you care about is the SIGN of the number. If the sign is
positive then you know the graph goes up on that side of the asymptote if it is negative it
goes down.
f ( x)
+
+
x
1.9
2.1
2.9
3.1
Graph goes up on the left side of x=2
Graph goes down on the right side of x=2
Graph goes down on the left side of x=3
Graph goes up on the right side of x=3
So we get something that looks like:
100
80
60
40
20
-2
0
-1 -20 0
-40
-60
-80
-100
-120
1
2
3
4
5
6
7
#21/p.382 Graph f ( x ) =
3 − 4x
x +1
Step 1: Find the asymptotes.
Vertical asymptote: Set denominator = 0 x + 1 = 0 ⇒ x = −1 is the vertical asymptote
Horizontal asymptote: Use the shortcut from #14 (case 2) to get y =
−4
= −4 as the
1
horizontal asymptote.
Next you want to see what happens to the function as x gets close to your asymptotes. To
do that you plug in numbers into f(x) that are close to -1. You need to try numbers on
both sides. The only thing you care about is the SIGN of the number. If the sign is
positive then you know the graph goes up on that side of the asymptote if it is negative it
goes down.
f ( x)
+
x
-1.1
-0.9
Graph goes down on the left side of x = -1
Graph goes up on the right side of x = -1
There is a horizontal asymptote here at y = -4 which is hard to see.
80
60
40
20
0
-5
-4
-3
-2
-1
0
-20
-40
-60
-80
-100
1
2
3
4
5
6
4 x 3 + 5x 2 + 6
#23/p.382 Find all asymptotes for f ( x ) = 2
.
x − 3x − 4
Vertical asymptote: Set denominator = 0 x 2 − 3x − 4 = ( x − 4)( x + 1) = 0
so x = 4 and x = -1 are the vertical asymptotes
There are no horizontal asymptotes because the numerator is a higher degree than the
denominator but there is an oblique asymptote.
To find the oblique asymptote you need to divide the two polynomials:
4 x + 17
x − 3x − 4 4 x + 5x + 0 x + 6 Remainder 67 x + 74
2
3
2
What this tells us is that the oblique asymptote is y = 4 x + 17 .
As a side note we also we now know that we can write:
4 x 3 + 5x 2 + 6
67 x + 74
= 4 x + 17 + 2
2
x − 3x − 4
x − 3x − 4