# 3.7 Rational functions  (Page 8/16)

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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 $\text{\hspace{0.17em}}f\left(x\right)=\frac{\left(x+2\right)\left(x-3\right)}{{\left(x+1\right)}^{2}\left(x-2\right)}.$

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:

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 $\text{\hspace{0.17em}}x=–2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=3.\text{\hspace{0.17em}}$ At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.

We have a y -intercept at $\text{\hspace{0.17em}}\left(0,3\right)\text{\hspace{0.17em}}$ and x -intercepts at $\text{\hspace{0.17em}}\left(–2,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,0\right).$

To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when $\text{\hspace{0.17em}}x+1=0\text{\hspace{0.17em}}$ and when $\text{\hspace{0.17em}}x–2=0,\text{\hspace{0.17em}}$ giving us vertical asymptotes at $\text{\hspace{0.17em}}x=–1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=2.$

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 $\text{\hspace{0.17em}}y=0.$

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] .

The factor associated with the vertical asymptote at $\text{\hspace{0.17em}}x=-1\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ 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.

Given the function $\text{\hspace{0.17em}}f\left(x\right)=\frac{{\left(x+2\right)}^{2}\left(x-2\right)}{2{\left(x-1\right)}^{2}\left(x-3\right)},\text{\hspace{0.17em}}$ use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

Horizontal asymptote at $\text{\hspace{0.17em}}y=\frac{1}{2}.\text{\hspace{0.17em}}$ Vertical asymptotes at y -intercept at $\text{\hspace{0.17em}}\left(0,\frac{4}{3}.\right)$

x -intercepts at $\left(–2,0\right)\text{\hspace{0.17em}}$ is a zero with multiplicity 2, and the graph bounces off the x -axis at this point. $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ is a single zero and the graph crosses the axis at this point. what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil By By By Rohini Ajay By Anh Dao By Stephen Voron By OpenStax By OpenStax By Marion Cabalfin By OpenStax By David Corey By Hope Percle By Monty Hartfield