# 3.7 Rational functions  (Page 9/16)

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## Writing rational functions

Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x -intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x -intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.

## Writing rational functions from intercepts and asymptotes

If a rational function    has x -intercepts at vertical asymptotes at $\text{\hspace{0.17em}}x={v}_{1},{v}_{2},\dots ,{v}_{m},\text{\hspace{0.17em}}$ and no then the function can be written in the form:

$f\left(x\right)=a\frac{{\left(x-{x}_{1}\right)}^{{p}_{1}}{\left(x-{x}_{2}\right)}^{{p}_{2}}\cdots {\left(x-{x}_{n}\right)}^{{p}_{n}}}{{\left(x-{v}_{1}\right)}^{{q}_{1}}{\left(x-{v}_{2}\right)}^{{q}_{2}}\cdots {\left(x-{v}_{m}\right)}^{{q}_{n}}}$

where the powers $\text{\hspace{0.17em}}{p}_{i}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}{q}_{i}\text{\hspace{0.17em}}$ on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ can be determined given a value of the function other than the x -intercept or by the horizontal asymptote if it is nonzero.

Given a graph of a rational function, write the function.

1. Determine the factors of the numerator. Examine the behavior of the graph at the x -intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)
2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.
3. Use any clear point on the graph to find the stretch factor.

## Writing a rational function from intercepts and asymptotes

Write an equation for the rational function shown in [link] .

The graph appears to have 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 both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at $\text{\hspace{0.17em}}x=–1\text{\hspace{0.17em}}$ seems to exhibit the basic behavior similar to $\text{\hspace{0.17em}}\frac{1}{x},\text{\hspace{0.17em}}$ with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ is exhibiting a behavior similar to $\text{\hspace{0.17em}}\frac{1}{{x}^{2}},\text{\hspace{0.17em}}$ with the graph heading toward negative infinity on both sides of the asymptote. See [link] .

We can use this information to write a function of the form

$f\left(x\right)=a\frac{\left(x+2\right)\left(x-3\right)}{\left(x+1\right){\left(x-2\right)}^{2}}.$

To find the stretch factor, we can use another clear point on the graph, such as the y -intercept $\text{\hspace{0.17em}}\left(0,–2\right).$

This gives us a final function of $\text{\hspace{0.17em}}f\left(x\right)=\frac{4\left(x+2\right)\left(x-3\right)}{3\left(x+1\right){\left(x-2\right)}^{2}}.$

Access these online resources for additional instruction and practice with rational functions.

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
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas