# 3.7 Rational functions

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In this section, you will:
• Use arrow notation.
• Solve applied problems involving rational functions.
• Find the domains of rational functions.
• Identify vertical asymptotes.
• Identify horizontal asymptotes.
• Graph rational functions.

Suppose we know that the cost of making a product is dependent on the number of items, $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ produced. This is given by the equation $\text{\hspace{0.17em}}C\left(x\right)=15,000x-0.1{x}^{2}+1000.\text{\hspace{0.17em}}$ If we want to know the average cost for producing $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ items, we would divide the cost function by the number of items, $\text{\hspace{0.17em}}x.$

The average cost function, which yields the average cost per item for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ items produced, is

$f\left(x\right)=\frac{15,000x-0.1{x}^{2}+1000}{x}$

Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.

In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.

## Using arrow notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in [link] , and notice some of their features.

Several things are apparent if we examine the graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}.$

1. On the left branch of the graph, the curve approaches the x -axis
2. As the graph approaches $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ from the left, the curve drops, but as we approach zero from the right, the curve rises.
3. Finally, on the right branch of the graph, the curves approaches the x- axis

To summarize, we use arrow notation    to show that $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is approaching a particular value. See [link] .

Arrow notation
Symbol Meaning
$x\to {a}^{-}$ $x\text{\hspace{0.17em}}$ approaches $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ from the left ( $x but close to $\text{\hspace{0.17em}}a$ )
$x\to {a}^{+}$ $x\text{\hspace{0.17em}}$ approaches $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ from the right ( $x>a\text{\hspace{0.17em}}$ but close to $\text{\hspace{0.17em}}a$ )
$x\to \infty$ $x\text{\hspace{0.17em}}$ approaches infinity ( $x\text{\hspace{0.17em}}$ increases without bound)
$x\to -\infty$ $x\text{\hspace{0.17em}}$ approaches negative infinity ( $x\text{\hspace{0.17em}}$ decreases without bound)
$f\left(x\right)\to \infty$ the output approaches infinity (the output increases without bound)
$f\left(x\right)\to -\infty$ the output approaches negative infinity (the output decreases without bound)
$f\left(x\right)\to a$ the output approaches $\text{\hspace{0.17em}}a$

## Local behavior of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}$

Let’s begin by looking at the reciprocal function, $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}.\text{\hspace{0.17em}}$ We cannot divide by zero, which means the function is undefined at $\text{\hspace{0.17em}}x=0;\text{\hspace{0.17em}}$ so zero is not in the domain . As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in [link] .

 $x$ –0.1 –0.01 –0.001 –0.0001 $f\left(x\right)=\frac{1}{x}$ –10 –100 –1000 –10,000

We write in arrow notation

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in [link] .

 $x$ 0.1 0.01 0.001 0.0001 $f\left(x\right)=\frac{1}{x}$ 10 100 1000 10,000

We write in arrow notation

This behavior creates a vertical asymptote , which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ as the input becomes close to zero. See [link] .

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
"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