In a rational function, the real zeros of the function are the same as the real zeros of the numerator (as long as these values are within the function's domain).

We can find the zeros of a rational function as follows:

Always start by factoring the numerator and denominator
Look for any common factors between the numerator and denominator and divide them out
Set what is left of the numerator equal to 0 and solve

The zeros of the denominator also have an important meaning. The zeros of both polynomials that make up the rational function r\left(x\right)are endpoints, or asymptotes, for the intervals satisfying the inequalites r\left(x\right) \geq 0 or r\left(x\right) \leq 0. We can visualize this on a graph.

-4

-3

-2

-1

-4

-3

-2

-1

r(x)

The function r\left(x\right)=\dfrac{\left(x+1\right)\left(x-2\right)}{x} is graphed.

The zeros of the polynomial in the numerator are: x=-1 and x=2 which correspond with the x-intercepts on the graph.

The zero of the polynomial in the denominator is: x=0 which corresponds with a vertical asymptote on the graph.

r\left(x\right) \geq 0 is shown by the green portion of the graph and r\left(x\right) \leq 0 is shown by the blue portion of the graph.

Examples

Example 1

Find the real zeros of r\left(x\right)=\dfrac{x^2 - 5x + 6}{x-3}.

Worked Solution

Create a strategy

Factor the numerator and divide out any common factors. Then set the numerator equal to 0 and solve.

Apply the idea

Factoring the numerator we get:

r\left(x\right)=\dfrac{\left(x-3\right)\left(x-2\right)}{x-3}

\left(x-3\right) is a common factor of the numerator and denominator so it can be divided out. This leaves:

r\left(x\right)=x-2

Setting this equal to 0 we get:

x-2=0

Solving we get that x=2 is the real zero of the function.

Reflect and check

If we had not divided out the common factor of x-3 we might have mistakenly gotten x=3 as a zero of the function. However, x=3 is no a zero of the function because it is not in the domain of the function. We will explore why this is in a future lesson.

Example 2

State the intervals that satisfy the following inequalities:

\dfrac{x^2-4}{x+1} \gt 0

Worked Solution

Create a strategy

Find the real zeros of the numerator and denominator to identify the zeros and vertical asymptotes of the function. Then use test points to determine the behavior of the function around the zeros and asymptotes.

Apply the idea

Find the real zeros of the numerator:

Solve the equation x^2-4=0 by factoring the difference of squares to get \left(x-2\right)\left(x+2\right)=0.

Applying the zero product property gives the real zeros x=2 and x=-2.

Find the real zeros of the denominator:

Solving the equation x+1=0 gives the real zero x=-1, which is the location of a vertical asymptote.

The critical points are at x=-2, x=-1 and x=2.

Next we will test the intervals between those critical points to find which intervals satify the inequality by being \gt 0.

Interval \left(-\infty,-2\right): Choose x=-3. \dfrac{\left(-3\right)^2-4}{\left(-3\right)+1}=\dfrac{9-4}{-2}=-2.5\lt0
Interval \left(-2,-1\right): Choose x=-1.5. \dfrac{\left(-1.5\right)^2-4}{\left(-1.5\right)+1}=\dfrac{2.25-4}{-0.5}=3.5\gt0
Interval \left(-1,2\right): Choose x=0. \dfrac{\left(0\right)^2-4}{\left(0\right)+1}=\dfrac{-4}{1}=-4\lt0
Interval \left(2,\infty\right): Choose x=3. \dfrac{\left(3\right)^2-4}{\left(3\right)+1}=\dfrac{9-4}{4}=1.25\gt0

Combining the intervals that satisfy the inequality by giving a result greater than 0 we get: \left(-2,-1\right) \cup \left(2,\infty\right).

Idea summary

Always start by factoring the numerator and denominator
Look for any common factors between the numerator and denominator and divide them out
Set what is left of the numerator equal to 0 and solve

1.8 Rational functions and zeros

Introduction

Ideas

Zeros of rational functions

Examples

Example 1

Example 2

Outcomes

1.8.A

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