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# Horizontal Asymptotes and Intercepts

### Learning Outcomes

• Use the degree of the numerator and denominator of a rational function to determine what kind of horizontal (or slant) asymptote it will have.
• Calculate slant asymptotes.
• Determine the intercepts of a rational function in factored form.
While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.

### Recall end behavior of a graph

We have learned that knowing the leading coefficient and degree of a polynomial function is useful when predicting its end behavior. If the leading term is positive or negative, and has even or odd degree, it will tell us the toolkit function's graph behavior it will mimic: $f(x)=x^2, \quad f(x)=-x^2,\quad f(x)=x^3,\quad$ or $\quad f(x)=-x^3$. The same idea applies to the ratio of leading terms of a rational function. See below for the three cases to check when determining horizontal (or slant) asymptotes of a rational function.
There are three distinct outcomes when checking for horizontal asymptotes: Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at $y=0$.

Example: $f\left(x\right)=\dfrac{4x+2}{{x}^{2}+4x - 5}$

In this case the end behavior is $f\left(x\right)\approx \frac{4x}{{x}^{2}}=\frac{4}{x}$. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function $g\left(x\right)=\frac{4}{x}$, and the outputs will approach zero, resulting in a horizontal asymptote at $y=0$. Note that this graph crosses the horizontal asymptote.
Horizontal Asymptote $y=0$ when $f\left(x\right)=\dfrac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne{0}\text{ where degree of }p<\text{degree of q}$.
Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.

Example: $f\left(x\right)=\dfrac{3{x}^{2}-2x+1}{x - 1}$

In this case the end behavior is $f\left(x\right)\approx \frac{3{x}^{2}}{x}=3x$. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function $g\left(x\right)=3x$. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of $g\left(x\right)=3x$ looks like a diagonal line, and since $f$ will behave similarly to $g$, it will approach a line close to $y=3x$. This line is a slant asymptote.
Slant Asymptote when $f\left(x\right)=\dfrac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0$ where degree of $p>\text{ degree of }q\text{ by }1$.
To find the equation of the slant asymptote, divide $\dfrac{3{x}^{2}-2x+1}{x - 1}$. The quotient is $3x+1$, and the remainder is 2. The slant asymptote is the graph of the line $g\left(x\right)=3x+1$. Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at $y=\frac{{a}_{n}}{{b}_{n}}$, where ${a}_{n}$ and ${b}_{n}$ are the leading coefficients of $p\left(x\right)$ and $q\left(x\right)$ for $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0$.

Example: $f\left(x\right)=\dfrac{3{x}^{2}+2}{{x}^{2}+4x - 5}$

In this case the end behavior is $f\left(x\right)\approx \frac{3{x}^{2}}{{x}^{2}}=3$. This tells us that as the inputs grow large, this function will behave like the function $g\left(x\right)=3$, which is a horizontal line. As $x\to \pm \infty ,f\left(x\right)\to 3$, resulting in a horizontal asymptote at $y=3$. Note that this graph crosses the horizontal asymptote.

Horizontal Asymptote when $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0\text{ where degree of }p=\text{degree of }q$.
Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote. 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

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

with end behavior

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

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

As $x\to \pm \infty , f\left(x\right)\to \infty$

### Try it

You'll need to graph the rational function $f\left(x\right)\ =\ \dfrac{\left(4x^a+2\right)}{\left(x^b+2x-3\right)}$ using an online graphing calculator. Adjust the $a$ and the $b$ values to obtain graphs with the following horizontal asymptotes:
1. Find values of $a$ and $b$ that will give a graph with a slant asymptote.
2. Find values of $a$ and $b$ that will give a graph with a horizontal asymptote at $y = 0$
3. Find values of $a$ and $b$ that will give a graph with a horizontal asymptote that is the ratio of the leading coefficients of $f(x)$.

1. Find values of $a$ and $b$ that will give a graph with a slant asymptote.  For example $b = 3, a = 4$
2. Find values of $a$ and $b$ that will give a graph with a horizontal asymptote at $y = 0$.  For example, $b = 4, a = 2$
3. Find values of $a$ and $b$ that will give a graph with a horizontal asymptote that is the ratio of the leading coefficients of $f(x)$.For example, $b = 4, a = 4$

### A General Note: 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.
• Case 1: Degree of numerator is less than degree of denominator: horizontal asymptote at $y=0$
• Case 2: Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
• If the degree of the numerator is greater than the degree of the denominator by more than one, the end behavior of the function's graph will mimic that of the graph of the reduced ratio of leading terms.
• Case 3: Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

### Example: Identifying Horizontal and Slant Asymptotes

For the functions below, identify the horizontal or slant asymptote.
1. $g\left(x\right)=\dfrac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}$
2. $h\left(x\right)=\dfrac{{x}^{2}-4x+1}{x+2}$
3. $k\left(x\right)=\dfrac{{x}^{2}+4x}{{x}^{3}-8}$

Answer: For these solutions, we will use $f\left(x\right)=\dfrac{p\left(x\right)}{q\left(x\right)}, q\left(x\right)\ne 0$.

1. $g\left(x\right)=\dfrac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}$: The degree of $p$ and the degree of $q$ are both equal to 3, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at $y=\frac{6}{2}$ or $y=3$.
2. $h\left(x\right)=\dfrac{{x}^{2}-4x+1}{x+2}$: The degree of $p=2$ and degree of $q=1$. Since $p>q$ by 1, there is a slant asymptote found at $\dfrac{{x}^{2}-4x+1}{x+2}$.The quotient is $x - 6$ and the remainder is 13. There is a slant asymptote at $y=-x - 6$.
3. $k\left(x\right)=\dfrac{{x}^{2}+4x}{{x}^{3}-8}$: The degree of $p=2\text{ }<$ degree of $q=3$, so there is a horizontal asymptote $y=0$.

### Try it

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Watch this video to see more worked examples of determining which kind of horizontal asymptote a rational function will have. https://youtu.be/A1tApZSE8nI

### Example: Identifying Horizontal Asymptotes

In the sugar concentration problem earlier, we created the equation $C\left(t\right)=\dfrac{5+t}{100+10t}$. Find the horizontal asymptote and interpret it in context of the problem.

Answer: 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 $t$, with coefficient 1. In the denominator, the leading term is $10t$, with coefficient 10. The horizontal asymptote will be at the ratio of these values:

$t\to \infty , C\left(t\right)\to \frac{1}{10}$

This function will have a horizontal asymptote at $y=\frac{1}{10}$. This tells us that as the values of $t$ increase, the values of $C$ will approach $\frac{1}{10}$. 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 $\frac{1}{10}$ pounds per gallon.

### Example: Identifying Horizontal and Vertical Asymptotes

Find the horizontal and vertical asymptotes of the function

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

Answer: 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 $x=1,-2,\text{and }5$, 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 $x\to \pm \infty , f\left(x\right)\to 0$. This function will have a horizontal asymptote at $y=0$.

### Try It

Find the vertical and horizontal asymptotes of the function:

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

Answer: Vertical asymptotes at $x=2$ and $x=-3$; horizontal asymptote at $y=4$.

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## Finding Intercepts of Rational Functions

### finding intercepts of a graph

Recall that we may always find any horizontal intercepts of the graph of a function by setting the output equal to zero. And we may always find any vertical intercept by setting the input equal to zero. In a rational function of the form $r(x)=\dfrac{P(x)}{Q(x)}$,
• Find the vertical intercept (the y-intercept) by evaluating $r(0)$. That is, replace all the input variables with $0$ and calculate the result.
• Find the horizontal intercept(s) (the x-intercepts) by solving $r(x)=0$. Since the function is undefined where the denominator equals zero], set the numerator equal to zero to find the horizontal intercepts of the function.
• Note that the graph of a rational function may not possess a vertical- or horizontal-intercept.

### A General Note: 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.

### Example: Finding the Intercepts of a Rational Function

Find the intercepts of $f\left(x\right)=\dfrac{\left(x - 2\right)\left(x+3\right)}{\left(x - 1\right)\left(x+2\right)\left(x - 5\right)}$.

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

\begin{align}f\left(0\right)&=\dfrac{\left(0 - 2\right)\left(0+3\right)}{\left(0 - 1\right)\left(0+2\right)\left(0 - 5\right)} \\[1mm] &=\frac{-6}{10} \\[1mm] &=-\frac{3}{5} \\[1mm] &=-0.6 \end{align}

The $x$-intercepts will occur when the function is equal to zero. A rational function is equal to 0 when the numerator is 0, as long as the denominator is not also 0.

\begin{align} 0&=\frac{\left(x - 2\right)\left(x+3\right)}{\left(x - 1\right)\left(x+2\right)\left(x - 5\right)} \\[1mm] 0&=\left(x - 2\right)\left(x+3\right)\end{align}

$x=2, -3$

The $y$-intercept is $\left(0,-0.6\right)$, the $x$-intercepts are $\left(2,0\right)$ and $\left(-3,0\right)$.

### Try It

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.

Answer: For the transformed reciprocal squared function, we find the rational form.

\begin{align}f\left(x\right)&=\dfrac{1}{{\left(x - 3\right)}^{2}}-4 \\[1mm] &=\dfrac{1 - 4{\left(x - 3\right)}^{2}}{{\left(x - 3\right)}^{2}} \\[1mm] &=\dfrac{1 - 4\left({x}^{2}-6x+9\right)}{\left(x - 3\right)\left(x - 3\right)} \\[1mm] &=\frac{-4{x}^{2}+24x - 35}{{x}^{2}-6x+9}\end{align}

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

Watch the following video to see more worked examples of finding asymptotes, intercepts and holes of rational functions. https://youtu.be/UnVZs2EaEjI