# 5.6 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] .

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] .

Need help solving this problem (2/7)^-2
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
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salma
Commplementary angles
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Sherica
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salma
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Ali
greetings from Iran
Ali
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hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_