A vertical asymptote is a vertical line, x=a, that the graph of a function approaches but never touches. In the context of rational functions, vertical asymptotes occur when the denominator of the function equals zero (provided the numerator does not also equal zero at the same value). This is because, as the denominator approaches zero, the value of the function increases or decreases without bound.

The multiplicity of a real zero in the denominator also plays a key role. If the multiplicity of the zero in the denominator is greater than its multiplicity in the numerator, a vertical asymptote occurs.

-2

f(x)

Multiplicity of x=2 is greater in numerator. No asymptote.

-2

Multiplicity of x=2 is greater in denominator. Vertical asymptote at x=2.

We can describe the behavior of a function around a vertical asymptote by looking at how the function changes as it approaches the asymptote from the left and from the right.

We can write this in limit notation as: \lim_{x \to a+} r\left(x\right)=\infty or \lim_{x \to a+} r\left(x\right)=-\infty

In words that means, "the limit of r\left(x\right) as x approaches a from the right is positive (or negative) infinity."

From the left side the limit notation is: \lim_{x \to a-} r\left(x\right)=\infty or \lim_{x \to a-} r\left(x\right)=-\infty

In words that means, "the limit of r\left(x\right) as x approaches a from the left is positive (or negative) infinity."

Examples

Example 1

Consider the function: s\left(x\right)=\dfrac{x^2-4}{x^2-2x}

Determine the vertical asymptotes of s\left(x\right).

Worked Solution

Create a strategy

Set the denominator equal to 0 and solve.

Apply the idea

Setting the denominator equal to zero we get the equation x^2-2x=0, which we can solve by factoring out the greatest common factor x.

After factoring we have: x\left(x-2\right)=0.

Using the zero product property we can find that x=0 or x-2=0.

Solving we get x=0 or x=2.

Before we can say that there is a vertical asymptote at each of these values we need to check that they are not also zeros of the numerator.

One way to check is by finding the zeros of the numerator which we can do by factoring.

x^2-4=\left(x+2\right)\left(x-2\right)

Setting the factors equal to zero we get x+2=0 and x-2=0.

Solving those equations gives us x=-2 and x=2.

We can see that x=2 was a zero in both the numerator and denominator (each with a multiplicity of 1) so there is no vertical asymptote at x=2 but there is a vertical asymptote at x=0 because that was only a zero of the denominator.

Reflect and check

Even though there is not a vertical asymptote at x=2 there is still something special happening there that we will learn about in the next lesson.

Describe the behavior of s\left(x\right) as it approaches any vertical asymptotes.

Worked Solution

Create a strategy

In part (a) we found that s\left(x\right) has a vertical asymptote at x=0. To describe how the function behaves near this asymptote we can examine the limits from both directions.

Apply the idea

Find the limit of the function as x approaches 0 from the left:

To do this we will substitute values of x into the function that are smaller, but very close to 0 intos\left(x\right).

\displaystyle s\left(-1\right)	\displaystyle =	\displaystyle \dfrac{\left(-1\right)^2-4}{\left(-1\right)^2-2\left(-1\right)}	Substitute x=-1
\displaystyle s\left(-1\right)	\displaystyle =	\displaystyle -1	Simplify

Next substitute a number even closer to 0.

\displaystyle s\left(-0.5\right)	\displaystyle =	\displaystyle \dfrac{\left(-0.5\right)^2-4}{\left(-0.5\right)^2-2\left(-0.5\right)}	Substitute x=-0.5
\displaystyle s\left(-0.5\right)	\displaystyle =	\displaystyle -3	Simplify

Finally substitute one more number even closer to 0.

\displaystyle s\left(-0.25\right)	\displaystyle =	\displaystyle \dfrac{\left(-0.25\right)^2-4}{\left(-0.25\right)^2-2\left(-0.25\right)}	Substitute x=-0.25
\displaystyle s\left(-0.25\right)	\displaystyle =	\displaystyle -7	Simplify

We can see the values are getting more negative as we approach the asymptote from the left. So we can say the value of the function is approaching -\infty as x approaches 0 from the left.

In limit notation that is: \lim_{x \to 0-} s\left(x\right)=-\infty

Taking the same approach from the right side, we can substitute values slightly larger than 0 to see what happens to the function values.

s\left(1\right)=3, \text{ } s\left(0.5\right)=5,\text{ }s\left(0.25\right)=9

As we approach the asymptote from the right the function values are getting larger. So we can say the value of the function is approaching \infty as x approaches 0 from the right.

In limit notation that is: \lim_{x \to 0+} s\left(x\right)=\infty

Example 2

Given the mathematical notation:

\lim_{x→2+} v\left(x\right) = \infty \\ \lim_{x→2-} v\left(x\right) = −\infty

Interpret the behavior of the rational function v\left(x\right) near the vertical asymptote x=2.

Worked Solution

Apply the idea

The limits describe the behavior of v\left(x\right) as it approaches x=2.

The limit \lim_{x→2+} v\left(x\right) = \infty has a + sign which means it is describing the function approaching from the right side. As the x-values approach 2 from the right, v\left(x\right) approaches \infty.

The limit \lim_{x→2-} v\left(x\right) = -\infty has a - sign which means it is describing the function approaching from the left side. As the x-values approach 2 from the right, v\left(x\right) approaches -\infty.

Idea summary

Vertical asymptotes occur in rational functions when the denominator equals zero and the numerator does not, or when the multiplicity of a zero in the denominator is greater than its multiplicity in the numerator.

When describing the behavior of a function around a vertical asymptote we use limit notation.

Approaching from the right: \lim_{x \to a+} r\left(x\right)=\infty or \lim_{x \to a+} r\left(x\right)=-\infty

Approaching from the left: \lim_{x \to a-} r\left(x\right)=\infty or \lim_{x \to a-} r\left(x\right)=-\infty

1.9 Rational functions and vertical asymptotes

Introduction

Ideas

Vertical asymptotes of rational functions

Examples

Example 1

Example 2

Outcomes

1.9.A

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