# 3.7 Rational functions  (Page 6/16)

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It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior    of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function

$f\left(x\right)=\frac{3{x}^{5}-{x}^{2}}{x+3}$

with end behavior

$f\left(x\right)\approx \frac{3{x}^{5}}{x}=3{x}^{4},$

the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.

## Horizontal asymptotes of rational functions

The horizontal asymptote    of a rational function can be determined by looking at the degrees of the numerator and denominator.

• Degree of numerator is less than degree of denominator: horizontal asymptote at $\text{\hspace{0.17em}}y=0.$
• Degree of numerator is greater than degree of denominator by one : no horizontal asymptote; slant asymptote.
• Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

## Identifying horizontal and slant asymptotes

For the functions below, identify the horizontal or slant asymptote.

1. $g\left(x\right)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}$
2. $h\left(x\right)=\frac{{x}^{2}-4x+1}{x+2}$
3. $k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}$

For these solutions, we will use

1. $g\left(x\right)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}:\text{\hspace{0.17em}}$ The degree of so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at $\text{\hspace{0.17em}}y=\frac{6}{2}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=3.$
2. $h\left(x\right)=\frac{{x}^{2}-4x+1}{x+2}:\text{\hspace{0.17em}}$ The degree of $\text{\hspace{0.17em}}p=2\text{\hspace{0.17em}}$ and degree of $\text{\hspace{0.17em}}q=1.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}p>q\text{\hspace{0.17em}}$ by 1, there is a slant asymptote found at $\text{\hspace{0.17em}}\frac{{x}^{2}-4x+1}{x+2}.$

The quotient is $\text{\hspace{0.17em}}x–6\text{\hspace{0.17em}}$ and the remainder is 13. There is a slant asymptote at $\text{\hspace{0.17em}}y=x–6.$

3. $k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}:\text{\hspace{0.17em}}$ The degree of degree of $\text{\hspace{0.17em}}q=3,\text{\hspace{0.17em}}$ so there is a horizontal asymptote $\text{\hspace{0.17em}}y=0.$

## Identifying horizontal asymptotes

In the sugar concentration problem earlier, we created the equation $\text{\hspace{0.17em}}C\left(t\right)=\frac{5+t}{100+10t}.$

Find the horizontal asymptote and interpret it in context of the problem.

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ with coefficient 1. In the denominator, the leading term is $\text{\hspace{0.17em}}10t,\text{\hspace{0.17em}}$ with coefficient 10. The horizontal asymptote will be at the ratio of these values:

This function will have a horizontal asymptote at $\text{\hspace{0.17em}}y=\frac{1}{10}.$

This tells us that as the values of t increase, the values of $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ will approach $\text{\hspace{0.17em}}\frac{1}{10}.\text{\hspace{0.17em}}$ In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or $\text{\hspace{0.17em}}\frac{1}{10}\text{\hspace{0.17em}}$ pounds per gallon.

## Identifying horizontal and vertical asymptotes

Find the horizontal and vertical asymptotes of the function

$f\left(x\right)=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-1\right)\left(x+2\right)\left(x-5\right)}$

First, note that this function has no common factors, so there are no potential removable discontinuities.

The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at indicating vertical asymptotes at these values.

The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as This function will have a horizontal asymptote at $\text{\hspace{0.17em}}y=0.\text{\hspace{0.17em}}$ See [link] .

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