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Precalculus : Polynomial Functions

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Example Questions

Example Question #7 : Find A Point Of Discontinuity

Find a point of discontinuity for the following function:

$s(x)=\frac{180-5x^2}{x^2-36}$

Possible Answers:

(-6, -5)

(-5, -6)

There are no points of discontinuity for this function.

(6, 5)

Correct answer:

(-6, -5)

Explanation:

Start by factoring the numerator and denominator of the function.

$s(x)=\frac{180-5x^2}{x^2-36}=\frac{-5(x^2-36)}{(x-6)(x+6)}=\frac{-5(x-6)(x+6)}{(x-6)(x+6)}$

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

Since $x=\pm 6$ is a zero for both the numerator and denominator, there is a point of discontinuity there. Since the final function is s(x)=-5 , (6, -5) and (-6, -5) are points of discontinuity.

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Example Question #8 : Find A Point Of Discontinuity

Find a point of discontinuity in the following function:

$f(x)=\frac{x^2+x-12}{x^2+2x-8}$

Possible Answers:

There is no point of discontinuity for this function.

(0, 4)

$\left(-2, \frac{5}{7}\right)$

$\left(-4, \frac{7}{6}\right)$

Correct answer:

$\left(-4, \frac{7}{6}\right)$

Explanation:

Start by factoring the numerator and denominator of the function.

$f(x)=\frac{x^2+x-12}{x^2+2x-8}=\frac{(x+4)(x-3)}{(x+4)(x-2)}=\frac{(x-3)}{(x-2)}$

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.

$\frac{-4-3}{-4-2}=\frac{7}{6}$

$\left(-4, \frac{7}{6}\right)$ is the point of discontinuity.

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Example Question #9 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$g(x)=\frac{x^2-1}{x^2+8x+7}$

Possible Answers:

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

There is no point of discontinuity.

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

(2, 4)

Correct answer:

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

Explanation:

Start by factoring the numerator and denominator of the function.

$g(x)=\frac{x^2-1}{x^2+8x+7}=\frac{(x-1)(x+1)}{(x+1)(x+7)}=\frac{x-1}{x+7}$

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.

$\frac{-1-1}{-1+7}=-\frac{1}{3}$

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

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Example Question #1 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$h(x)=\frac{x^2-5x+6}{x^2-9}$

Possible Answers:

There is no point of discontinuity for this function.

$\left(6, -\frac{2}{3}\right)$

$\left(5, \frac{1}{8}\right)$

$\left(3, \frac{1}{6}\right)$

Correct answer:

$\left(3, \frac{1}{6}\right)$

Explanation:

Start by factoring the numerator and denominator of the function.

$h(x)=\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3}$

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.

$\frac{3-2}{3+3}=\frac{1}{6}$

$\left(3, \frac{1}{6}\right)$ 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:

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

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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Precalculus : Polynomial Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #7 : Find A Point Of Discontinuity

Example Question #8 : Find A Point Of Discontinuity

Example Question #9 : Find A Point Of Discontinuity

Example Question #1 : Find A Point Of Discontinuity

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

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