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Polynomials and Rational Functions
Rational Functions
Algebra Textbooks Boundless Algebra Polynomials and Rational Functions Rational Functions
Algebra Textbooks Boundless Algebra Polynomials and Rational Functions
Algebra Textbooks Boundless Algebra
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Algebra
Concept Version 10
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Solving Problems with Rational Functions

The xxx-intercepts of rational functions are found by setting the polynomial in the numerator equal to 000 and solving for xxx.

Learning Objective

  • Use the numerator of a rational function to solve for its zeros


Key Points

    • The xxx-intercepts (also known as zeros or roots) of a function are points where the graph intersects the xxx-axis. Rational functions can have zero, one, or multiple xxx-intercepts. 
    • For any function, the xxx-intercepts are xxx-values for which the function has a value of zero: f(x)=0f(x) = 0f(x)=0.
    • For rational functions, the xxx-intercepts exist when the numerator is equal to 000. For f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)}f(x)=​Q(x)​​P(x)​​, if P(x)=0P(x) = 0P(x)=0, then f(x)=0f(x) = 0f(x)=0.

Terms

  • denominator

    The number or expression written below the line in a fraction (thus 222 in 12\frac {1}{2}​2​​1​​).

  • rational function

    Any function whose value can be expressed as the quotient of two polynomials (except division by zero).

  • numerator

    The number or expression written above the line in a fraction (thus 111 in 12\frac {1}{2}​2​​1​​).


Full Text

Finding the xxx-intercepts of Rational Functions

Recall that a rational function is defined as the ratio of two real polynomials with the condition that the polynomial in the denominator is not a zero polynomial.

f(x)=P(x)Q(x)f(x) = \dfrac{P(x)}{Q(x)}f(x)=​Q(x)​​P(x)​​, where Q(x)≠0Q(x) \neq 0Q(x)≠0

An example of a rational function is:

f(x)=x+12x2−x−1f(x) = \dfrac{x + 1}{2x^2 - x - 1}f(x)=​2x​2​​−x−1​​x+1​​

Rational functions can be graphed on the coordinate plane. We can use algebraic methods to calculate their xxx-intercepts (also known as zeros or roots), which are points where the graph intersects the xxx-axis. Rational functions can have zero, one, or multiple xxx-intercepts. 

For any function, the xxx-intercepts are xxx-values for which the function has a value of zero: f(x)=0f(x) = 0f(x)=0.

In the case of rational functions, the xxx-intercepts exist when the numerator is equal to 000. For f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)}f(x)=​Q(x)​​P(x)​​, if P(x)=0P(x) = 0P(x)=0, then f(x)=0f(x) = 0f(x)=0.

In order to solve rational functions for their xxx-intercepts, set the polynomial in the numerator equal to zero, and solve for xxx by factoring where applicable.

Example 1

Find the xxx-intercepts of this function:

f(x)=x2−3x+2x2−2x−3f(x) = \dfrac{x^2 - 3x + 2}{x^2 - 2x -3}f(x)=​x​2​​−2x−3​​x​2​​−3x+2​​ 

Set the numerator of this rational function equal to zero and solve for xxx: 

0=x2−3x+2=(x−1)(x−2)\begin {aligned} 0 &=x^2 - 3x + 2 \\&= (x - 1)(x - 2) \end {aligned}​0​​​​=x​2​​−3x+2​=(x−1)(x−2)​​

Solutions for this polynomial are x=1x = 1x=1 or x=2x= 2x=2. This means that this function has xxx-intercepts at 111 and 222.

Example 2

Find the xxx-intercepts of the function:

f(x)=1xf(x) = \dfrac {1}{x}f(x)=​x​​1​​ 

Here, the numerator is a constant, and therefore, cannot be set equal to 000. Thus, this function does not have any xxx-intercepts.

Example 3

Find the roots of:

 g(x)=x3−2x2x2−10g(x) = \dfrac{x^3 - 2x}{2x^2 - 10} g(x)=​2x​2​​−10​​x​3​​−2x​​

Factoring the numerator, we have:

 0=x3−2x=x(x2−2)\begin {aligned} 0&=x^3 - 2x \\&= x(x^2 - 2) \end {aligned}​0​​​​=x​3​​−2x​=x(x​2​​−2)​​

Given the factor xxx, the polynomial equals 000 when x=0x=0x=0. 

Let the second factor equal zero, and solve for xxx:

$x^2 - 2 = 0 \\ x^2 = 2 \\ x = \pm \sqrt{2}$

Thus there are three roots, or xxx-intercepts: 000, −2-\sqrt{2}−√​2​​​ and 2\sqrt{2}√​2​​​. These can be observed in the graph of the function below.

Graph of g(x)=x3−2x2x2−10g(x) = \frac{x^3 - 2x}{2x^2 - 10}g(x)=​2x​2​​−10​​x​3​​−2x​​

xxx-intercepts exist at x=−2,0,2x = -\sqrt{2}, 0, \sqrt{2}x=−√​2​​​,0,√​2​​​.

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