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Find the vertical and horizontal asymptotes of the function:

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

Vertical asymptotes at x = 2 and x = 3 ; horizontal asymptote at y = 4.

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Intercepts of rational functions

A rational function    will have a y -intercept when the input is zero, if the function is defined at zero. A rational function will not have a y -intercept if the function is not defined at zero.

Likewise, a rational function will have x -intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x -intercepts can only occur when the numerator of the rational function is equal to zero.

Finding the intercepts of a rational function

Find the intercepts of f ( x ) = ( x 2 ) ( x + 3 ) ( x 1 ) ( x + 2 ) ( x 5 ) .

We can find the y -intercept by evaluating the function at zero

f ( 0 ) = ( 0 2 ) ( 0 + 3 ) ( 0 1 ) ( 0 + 2 ) ( 0 5 )          = 6 10          = 3 5         = 0.6

The x -intercepts will occur when the function is equal to zero:

0 = ( x 2 ) ( x + 3 ) ( x 1 ) ( x + 2 ) ( x 5 ) This is zero when the numerator is zero . 0 = ( x 2 ) ( x + 3 ) x = 2 ,   3

The y -intercept is ( 0 , –0.6 ) , the x -intercepts are ( 2 , 0 ) and ( –3 , 0 ) . See [link] .

Graph of f(x)=(x-2)(x+3)/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5, its horizontal asymptote at y=0, and its intercepts at (-3, 0), (0, -0.6), and (2, 0).
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Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x - and y -intercepts and the horizontal and vertical asymptotes.

For the transformed reciprocal squared function, we find the rational form. f ( x ) = 1 ( x 3 ) 2 4 = 1 4 ( x 3 ) 2 ( x 3 ) 2 = 1 4 ( x 2 6 x + 9 ) ( x 3 ) ( x 3 ) = 4 x 2 + 24 x 35 x 2 6 x + 9

Because the numerator is the same degree as the denominator we know that as x ± ,   f ( x ) 4 ;   so   y = 4 is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is x = 3 , because as x 3 , f ( x ) . We then set the numerator equal to 0 and find the x -intercepts are at ( 2.5 , 0 ) and ( 3.5 , 0 ) . Finally, we evaluate the function at 0 and find the y -intercept to be at ( 0 , 35 9 ) .

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Graphing rational functions

In [link] , we see that the numerator of a rational function reveals the x -intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See [link] .

Graph of y=1/x with its vertical asymptote at x=0.

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See [link] .

Graph of y=1/x^2 with its vertical asymptote at x=0.

For example, the graph of f ( x ) = ( x + 1 ) 2 ( x 3 ) ( x + 3 ) 2 ( x 2 ) is shown in [link] .

Graph of f(x)=(x+1)^2(x-3)/(x+3)^2(x-2) with its vertical asymptotes at x=-3 and x=2, its horizontal asymptote at y=1, and its intercepts at (-1, 0), (0, 1/6), and (3, 0).
  • At the x -intercept x = 1 corresponding to the ( x + 1 ) 2 factor of the numerator, the graph bounces, consistent with the quadratic nature of the factor.
  • At the x -intercept x = 3 corresponding to the ( x 3 ) factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
  • At the vertical asymptote x = 3 corresponding to the ( x + 3 ) 2 factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function f ( x ) = 1 x 2 .
  • At the vertical asymptote x = 2 , corresponding to the ( x 2 ) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function f ( x ) = 1 x .

Questions & Answers

foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
Churlene Reply
difference between calculus and pre calculus?
Asma Reply
give me an example of a problem so that I can practice answering
Jenefa Reply
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
CJ Reply
I want to learn about the law of exponent
Quera Reply
explain this
Hinderson Reply
what is functions?
Angel Reply
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?
Feemark Reply
can you not take the square root of a negative number
Sharon Reply
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
Elaine Reply
can I get some pretty basic questions
Ama Reply
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
Amara Reply
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
Mars Reply
what is the domain of f(x)=x-4/x^2-2x-15 then
Conney Reply
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
jeric Reply
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
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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