<< Chapter < Page Chapter >> Page >

Vertical asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

Given a rational function, identify any vertical asymptotes of its graph.

  1. Factor the numerator and denominator.
  2. Note any restrictions in the domain of the function.
  3. Reduce the expression by canceling common factors in the numerator and the denominator.
  4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
  5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.

Identifying vertical asymptotes

Find the vertical asymptotes of the graph of k ( x ) = 5 + 2 x 2 2 x x 2 .

First, factor the numerator and denominator.

k ( x ) = 5 + 2 x 2 2 x x 2         = 5 + 2 x 2 ( 2 + x ) ( 1 x )

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

( 2 + x ) ( 1 x ) = 0                      x = 2 , 1

Neither x = 2 nor x = 1 are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in [link] confirms the location of the two vertical asymptotes.

Graph of k(x)=(5+2x)^2/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.
Got questions? Get instant answers now!
Got questions? Get instant answers now!

Removable discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity    .

For example, the function f ( x ) = x 2 1 x 2 2 x 3 may be re-written by factoring the numerator and the denominator.

f ( x ) = ( x + 1 ) ( x 1 ) ( x + 1 ) ( x 3 )

Notice that x + 1 is a common factor to the numerator and the denominator. The zero of this factor, x = 1 , is the location of the removable discontinuity. Notice also that x 3 is not a factor in both the numerator and denominator. The zero of this factor, x = 3 , is the vertical asymptote. See [link] .

Graph of f(x)=(x^2-1)/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.

Removable discontinuities of rational functions

A removable discontinuity    occurs in the graph of a rational function at x = a if a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.

Identifying vertical asymptotes and removable discontinuities for a graph

Find the vertical asymptotes and removable discontinuities of the graph of k ( x ) = x 2 x 2 4 .

Factor the numerator and the denominator.

k ( x ) = x 2 ( x 2 ) ( x + 2 )

Notice that there is a common factor in the numerator and the denominator, x 2. The zero for this factor is x = 2. This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator, x + 2. The zero for this factor is x = 2. The vertical asymptote is x = 2. See [link] .

Graph of k(x)=(x-2)/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.

The graph of this function will have the vertical asymptote at x = −2 , but at x = 2 the graph will have a hole.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Questions & Answers

So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
KARMEL Reply
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
Rima Reply
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
Jhon Reply
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's $26.50 monthly payment, you'll need 3,000 texts which will cost an additional $10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
Practice Key Terms 5

Get the best Precalculus course in your pocket!





Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Precalculus' conversation and receive update notifications?

Ask