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Given a rational function, sketch a graph.

  1. Evaluate the function at 0 to find the y -intercept.
  2. Factor the numerator and denominator.
  3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x -intercepts.
  4. Find the multiplicities of the x -intercepts to determine the behavior of the graph at those points.
  5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.
  6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.
  7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
  8. Sketch the graph.

Graphing a rational function

Sketch a graph of f ( x ) = ( x + 2 ) ( x 3 ) ( x + 1 ) 2 ( x 2 ) .

We can start by noting that the function is already factored, saving us a step.

Next, we will find the intercepts. Evaluating the function at zero gives the y -intercept:

f ( 0 ) = ( 0 + 2 ) ( 0 3 ) ( 0 + 1 ) 2 ( 0 2 )         = 3

To find the x -intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x -intercepts at x = –2 and x = 3. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.

We have a y -intercept at ( 0 , 3 ) and x -intercepts at ( –2 , 0 ) and ( 3 , 0 ) .

To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when x + 1 = 0 and when x 2 = 0 , giving us vertical asymptotes at x = –1 and x = 2.

There are no common factors in the numerator and denominator. This means there are no removable discontinuities.

Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0.

To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x -intercepts between the vertical asymptotes, and the y -intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in [link] .

Graph of only the middle portion of f(x)=(x+2)(x-3)/(x+1)^2(x-2) with its intercepts at (-2, 0), (0, 3), and (3, 0).

The factor associated with the vertical asymptote at x = −1 was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.

For the vertical asymptote at x = 2 , the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See [link] . After passing through the x -intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.

Graph of f(x)=(x+2)(x-3)/(x+1)^2(x-2) with its vertical asymptotes at x=-1 and x=2, its horizontal asymptote at y=0, and its intercepts at (-2, 0), (0, 3), and (3, 0).
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Given the function f ( x ) = ( x + 2 ) 2 ( x 2 ) 2 ( x 1 ) 2 ( x 3 ) , use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

Horizontal asymptote at y = 1 2 . Vertical asymptotes at x = 1   and   x = 3. y -intercept at ( 0 , 4 3 . )

x -intercepts at ( 2 , 0 )    and  ( 2 , 0 ) . ( 2 , 0 ) is a zero with multiplicity 2, and the graph bounces off the x -axis at this point. ( 2 , 0 ) is a single zero and the graph crosses the axis at this point.

Graph of f(x)=(x+2)^2(x-2)/2(x-1)^2(x-3) with its vertical and horizontal asymptotes.
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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

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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