Precalculus Name: Period: ____ Date: Worksheet: Vertical

Precalculus
Name: _______________________
Period: ____
Date: _______________________
Worksheet: Vertical Asymptotes
We’re now going to start working with rational functions. A rational function is just a fraction of polynomials.
Example of a rational function:
𝑔(π‘₯) =
2π‘₯ 3 +π‘₯ 2 βˆ’5π‘₯+1
π‘₯ 4 +3π‘₯βˆ’6
To keep things simple, we’ll first only look at rational functions with a numerator of 1. For problems 1 through
𝟏
4, refer to the function 𝒇(𝒙) = π’™βˆ’πŸ‘.
1. What is the parent function of 𝑓(π‘₯)?
2. How is 𝑓(π‘₯) shifted compared to the parent function?
3. When π‘₯ = 3, 𝑓(π‘₯) is undefined. We can only describe what happens to 𝑓(π‘₯) when π‘₯ is close to 3.
Complete the following table to find out what happens to the function on the left of π‘₯ = 3 (β€œas π‘₯ β†’ 3βˆ’ β€œ)
and on the right of π‘₯ = 3 (β€œas π‘₯ β†’ 3+ ”). You might want to use the table feature on your calculator.
What Happens as x Approaches 3 from the Left
What Happens as x Approaches 3 from the Right
Notation: As π‘₯ β†’ 3βˆ’
Notation: Notation: As π‘₯ β†’ 3+
𝑓(2) = _______
𝑓(4) = _______
𝑓(2.9) = _______
𝑓(3.1) = _______
𝑓(2.99) = _______
𝑓(3.01) = _______
𝑓(2.999) = _______
𝑓(3.001) = _______
𝑓(2.9999) = _______
𝑓(3.0001) = _______
βˆ’
Conclusion: As π‘₯ β†’ 3 , 𝑓(π‘₯) β†’_______.
Conclusion: As π‘₯ β†’ 3+ , 𝑓(π‘₯) β†’_______.
4. In problem (3), you should have found that as x approaches 3 from the left, 𝑓(π‘₯) approaches negative
infinity (as π‘₯ β†’ 3βˆ’ , 𝑓(π‘₯) β†’ βˆ’βˆž); and as x approaches 3 from the right, 𝑓(π‘₯) approaches positive infinity
(as π‘₯ β†’ 3+ , 𝑓(π‘₯) β†’ ∞). The graph of 𝑓(π‘₯) does not cross or touch the line π‘₯ = 3. It just gets closer and
closer to this line on either side. This feature is called an asymptote. And since it is a vertical line, it is
called a vertical asymptote. (Horizontal asymptotes also exist.) We always draw asymptotes with dotted
lines. On the axes below, graph the function 𝑓(π‘₯), showing the asymptote π‘₯ = 3 as a dotted line. Use the
table on your calculator to find the y values of other points with x values between –10 and 10.
5. What would be the equation for the vertical
1
asymptote in the graph of the function 𝑓(π‘₯) = π‘₯+5?
6. We get an interesting pattern when we raise the denominator to different powers. Sketch graphs of the
following functions. Fill in the blanks to describe whether the function goes up or down on each side of the
vertical asymptote.
1
1
𝑓(π‘₯) = (π‘₯βˆ’4)1
𝑓(π‘₯) = (π‘₯βˆ’4)2
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
1
𝑓(π‘₯) = (π‘₯βˆ’4)3
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
1
𝑓(π‘₯) = (π‘₯βˆ’4)4
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
1
𝑓(π‘₯) = (π‘₯βˆ’4)5
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
1
𝑓(π‘₯) = (π‘₯βˆ’4)6
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
7. Describe the pattern illustrated in problem (6).
8. Now show what happens when the whole function is multiplied by negative one by graphing the following
two functions:
1
1
𝑓(π‘₯) = βˆ’ (π‘₯βˆ’4)1
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
𝑓(π‘₯) = βˆ’ (π‘₯βˆ’4)2
As π‘₯ β†’ 4βˆ’ , 𝑓(π‘₯) β†’ _____.
As π‘₯ β†’ 4+ , 𝑓(π‘₯) β†’ _____.
9. A vertical asymptote occurs whenever the denominator of a rational function is equal to zero. It is possible
for the denominator to have more than one zero, so it is also possible for a rational function to have more
1
than one vertical asymptote. For example, the vertical asymptotes of the function 𝑓(π‘₯) = π‘₯ 2 βˆ’4 are π‘₯ = βˆ’2
1
and π‘₯ = 2, since π‘₯ 2 βˆ’ 4 factors to (π‘₯ βˆ’ 2)(π‘₯ + 2). Graph the function 𝑓(π‘₯) = π‘₯ 2 βˆ’4 on the axes below,
showing the vertical asymptotes as dotted lines. Use the table on your calculator to plot points. The graph
on your calculator’s screen will not look right because your calculator can’t handle vertical asymptotes.
10. Find the equations for the vertical asymptotes of the following rational functions:
1
1
𝑓(π‘₯) = π‘₯ 2 βˆ’9 :
𝑓(π‘₯) = βˆ’ π‘₯ 2 βˆ’2π‘₯βˆ’15 :
1
𝑓(π‘₯) = βˆ’ π‘₯ 3 βˆ’π‘₯ 2 βˆ’6π‘₯ :
𝑓(π‘₯) = π‘₯ 3 +π‘₯ 2 βˆ’4π‘₯βˆ’4 :
1
All of the functions we’ve looked at so far approach zero on both the left and the right. (As π‘₯ β†’ βˆ’βˆž, 𝑓(π‘₯) β†’
0; and as π‘₯ β†’ ∞, 𝑓(π‘₯) β†’ 0.) This means that all of these functions have a horizontal asymptote. Specifically,
𝑦 = 0. From now on, draw all horizontal asymptotes as dotted lines as well.
11. Graph the following functions, showing all vertical and horizontal asymptotes as dotted lines. Find the
vertical asymptotes by finding the zeros of the denominator. For each vertical asymptote, state the
behavior of the graph on either side. (For example, if the asymptote is π‘₯ = 2, state that as π‘₯ β†’ 2βˆ’ , 𝑓(π‘₯) β†’
±βˆž; and as π‘₯ β†’ 2+ , 𝑓(π‘₯) β†’ ±βˆž.)
1
𝑓(π‘₯) = π‘₯ 2 βˆ’16
Asymptote behavior:
Asymptote behavior:
1
𝑓(π‘₯) = π‘₯ 3 +π‘₯ 2 βˆ’2π‘₯βˆ’2
Asymptote behavior:
1
𝑓(π‘₯) = βˆ’ π‘₯ 2 +π‘₯βˆ’12
1
𝑓(π‘₯) = βˆ’ π‘₯ 4 βˆ’π‘₯ 3 βˆ’6π‘₯ 2
Asymptote behavior: