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Precalculus : Rational Functions

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Example Questions

Example Question #11 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$j(x)=\frac{x^2+7x+6}{x^2-1}$

Possible Answers:

$\left(-1, -\frac{5}{2}\right)$

(-3, 6)

$\left(1, \frac{5}{2}\right)$

There is no point of discontinuity for this function.

Correct answer:

$\left(-1, -\frac{5}{2}\right)$

Explanation:

Start by factoring the numerator and denominator of the function.

$j(x)=\frac{x^2+7x+6}{x^2-1}=\frac{(x+6)(x+1)}{(x+1)(x-1)}=\frac{x+6}{x-1}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since x=-1 is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{-1+6}{-1-1}=-\frac{5}{2}$

$\left(-1, -\frac{5}{2}\right)$ is the point of discontinuity.

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Example Question #11 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$k(x)=\frac{2x^2-x-1}{x^2-6x+5}$

Possible Answers:

$\left(1, -\frac{3}{4}\right)$

$\left(-5, \frac{9}{10}\right)$

There is no point of discontinuity for this function.

(1, 4)

Correct answer:

$\left(1, -\frac{3}{4}\right)$

Explanation:

Start by factoring the numerator and denominator of the function.

$k(x)=\frac{2x^2-x-1}{x^2-6x+5}=\frac{(2x+1)(x-1)}{(x-1)(x-5)}=\frac{2x+1}{x-5}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since x=1 is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{2(1)+1}{1-5}=-\frac{3}{4}$

$\left(1, -\frac{3}{4}\right)$ is the point of discontinuity.

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Example Question #21 : Rational Functions

Find the point of discontinuity for the following function:

$l(x)=\frac{2x^2+5x-12}{x^2+3x-4}$

Possible Answers:

$\left(-4, \frac{11}{5}\right)$

There is no point of discontinuity for this function.

(-3, 0)

(4, 1)

Correct answer:

$\left(-4, \frac{11}{5}\right)$

Explanation:

Start by factoring the numerator and denominator of the function.

$l(x)=\frac{2x^2+5x-12}{x^2+3x-4}=\frac{(2x-3)(x+4)}{(x+4)(x-1)}=\frac{2x-3}{x-1}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since x=-4 is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{2(-4)-3}{-4-1}=\frac{11}{5}$

$\left(-4, \frac{11}{5}\right)$ is the point of discontinuity.

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Example Question #12 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$n(x)=\frac{x^3+2x^2-3x}{x^2+8x+15}$

Possible Answers:

(-3, 6)

There is no point of discontinuity for this function.

(-4, 20)

$\left(-1, \frac{1}{2}\right)$

Correct answer:

(-3, 6)

Explanation:

Start by factoring the numerator and denominator of the function.

$n(x)=\frac{x^3+2x^2-3x}{x^2+8x+15}=\frac{x(x^2+2x-3)}{(x+3)(x+5)}=\frac{x(x-1)(x+3)}{(x+3)(x+5)}=\frac{x(x-1)}{x+5}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since x=-3 is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{-3(-3-1)}{-3+5}=6$

(-3, 6) is the point of discontinuity.

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Example Question #12 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$q(x)=\frac{x^2+x}{x^3+3x^2-x-3}$

Possible Answers:

$\left(-1, \frac{1}{4}\right)$

(-3, 2)

(0, 4)

There is no discontinuity for this function.

Correct answer:

$\left(-1, \frac{1}{4}\right)$

Explanation:

Start by factoring the numerator and denominator of the function.

$q(x)=\frac{x^2+x}{x^3+3x^2-x-3}=\frac{x(x+1)}{(3x^2-3)(x^3-x)}$

$=\frac{x(x+1)}{3(x^2-1)x(x^2-1)}=\frac{x(x+1)}{(x-1)(x+1)(x+3)}=\frac{x}{(x-1)(x+3)}$

A point of discontinuity occurs when a number is both a zero of the numerator and denominator.

Since x=-1 is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the value, plug in into the final simplified equation.

$\frac{-1}{(-1-1)(-1+3)}=\frac{1}{4}$

$\left(-1, \frac{1}{4}\right)$ is the point of discontinuity.

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Example Question #13 : Find A Point Of Discontinuity

Given the function, $y=\frac{8x-4}{4x-2}$ , where and what is the type of discontinuity, if any?

Possible Answers:

$\textup{Hole at }x=-\frac{1}{2} \textup{ and } x=\frac{1}{2}$

$\textup{Asymptote at }x=\frac{1}{2}$

$\textup{Hole and Asymptote at }x=\frac{1}{2}$

$\textup{There are no discontinuities.}$

$\textup{Hole at }x=\frac{1}{2}$

Correct answer:

$\textup{Hole at }x=\frac{1}{2}$

Explanation:

Before we simplify, set the denominator equal to zero to determine where is invalid. The value of the denominator cannot equal to zero.

$4x-2 \neq0$

$4x \neq2$

$x\neq \frac{2}{4} \neq \frac{1}{2}$

The value at $x=\frac{1}{2}$ is invalid in the domain.

Pull out a greatest common factor for the numerator and the denominator and simplify.

$y=\frac{8x-4}{4x-2} =\frac{4(2x-1)}{2(2x-1)} = 2$

Since the terms 2x-1 can be cancelled, there will not be any vertical asymptotes. Even though the rational function simplifies to y=2 , there will be instead a hole at $x=\frac{1}{2}$ on the graph.

The answer is: $\textup{Hole at }x=\frac{1}{2}$

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Example Question #14 : Find A Point Of Discontinuity

Find the point of discontinuity in the function $f(x)=\frac{4x^3-1}{x-4}$ .

Possible Answers:

x=6

x=0

x=4

$x=\frac{1}{4}$

x=-4

Correct answer:

x=4

Explanation:

When dealing with a rational expression, the point of discontinuity occurs when the denominator would equal 0. In this case, x-4=0 so x=4 . Therefore, your point of discontinuity is x=4 .

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Example Question #1 : Find Intercepts And Asymptotes

Suppose the function below has an oblique (i.e. slant asymptote) at y=3x .

$\frac{ax^n+3x+5}{bx^m+6}$

If we are given n>2 , what can we say about the relation between and and between and ?

Possible Answers:

$n = m-1\, \textup{and} \, \frac{b}{a}=3$

$n = m+1\, \textup{and} \, \frac{a}{b}=3$

$n = m+1\, \textup{and} \, \frac{b}{a}=3$

$n = m\, \textup{and} \, \frac{b}{a}=3$

$n = m-1\, \textup{and} \, \frac{a}{b}=3$

Correct answer:

$n = m+1\, \textup{and} \, \frac{a}{b}=3$

Explanation:

We can only have an oblique asymptote if the degree of the numerator is one more than the degree of the denominator. This stipulates that must equal m+1 .

The slope of the asymptote is determined by the ratio of the leading terms, which means the ratio of to must be 3 to 1. The actual numbers are not important.

Finally, since the value of is at least three, we know there is no intercept to our oblique asymptote.

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Example Question #1 : Find Intercepts And Asymptotes

Find the -intercept and asymptote, if possible.

$y=\frac{x^2+4x+3}{(x+3)}$

Possible Answers:

$\textup{Y-intercept at: } y=1; \textup{ Asymptote at: }x=3$

$\textup{Y-intercept at: } y=1; \textup{ Asymptote at: }x=-3$

$\textup{Y-intercept at: } y=1; \textup{ No asymptotes.}$

$\textup{Y-intercept at: } y=3; \textup{ No asymptotes.}$

$\textup{Y-intercept at: } y=3; \textup{ Asymptote at: }x=-3$

Correct answer:

$\textup{Y-intercept at: } y=1; \textup{ No asymptotes.}$

Explanation:

To find the y-intercept of $y=\frac{x^2+4x+3}{(x+3)}$ , simply substitute x=0 and solve for .

$y=\frac{0^2+4(0)+3}{(0+3)}=\frac{3}{3}=1$

The y-intercept is 1.

The numerator, x^2+4x+3 , can be simplified by factoring it into two binomials.

x^2+4x+3=(x+3)(x+1)

$y=\frac{x^2+4x+3}{(x+3)}=\frac{(x+3)(x+1)}{(x+3)} = x+1$

There is a removable discontinuity at x=-3 , but there are no asymptotes at x=-3 since the x+3 terms can be canceled.

The correct answer is: $\textup{Y-intercept at: } y=1; \textup{ No asymptotes.}$

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Example Question #2 : Find Intercepts And Asymptotes

Find the $\small x$ -intercepts of the rational function

$\small \frac{2x^2-6x+4}{2x^2+x-1}$ .

Possible Answers:

$\small x=1/2,-1$

$\small x=1$

$\small x=2,-1$

$\small x=1,2$

x=6,-2

Correct answer:

$\small x=1,2$

Explanation:

The $\small x$ -intercept(s) is/are the root(s) of the numerator of the rational functions.

In this case, the numerator is $\small 2x^2-6x+4$ .

Using the quadratic formula,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{6\pm\sqrt{(-6)^2-4(2)(4)}}{2(2)}$

$x=\frac{6\pm\sqrt{36-32}}{4}$

$x=\frac{6\pm2}{4}=1,2$

the roots are $\small x=1,2$ .

Thus, $\small x=1,2$ are the $\small x$ -intercepts.

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Precalculus : Rational Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Find A Point Of Discontinuity

Example Question #11 : Find A Point Of Discontinuity

Example Question #21 : Rational Functions

Example Question #12 : Find A Point Of Discontinuity

Example Question #12 : Find A Point Of Discontinuity

Example Question #13 : Find A Point Of Discontinuity

Example Question #14 : Find A Point Of Discontinuity

Example Question #1 : Find Intercepts And Asymptotes

Example Question #1 : Find Intercepts And Asymptotes

Example Question #2 : Find Intercepts And Asymptotes

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