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Precalculus : Find a Point of Discontinuity

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Example Question #1 : Find A Point Of Discontinuity

What are the holes or vertical asymptotes, if any, for the function: $y=\frac{x^2-1}{x-1}$

Possible Answers:

$\textup{Hole at } x=-1; \textup{ No vertical asymptotes.}$

$\textup{Hole at }x=1,-1$

$\textup{Hole at } x=1; \textup{ No vertical asymptotes.}$

$\textup{Hole at }x=1; \textup{Vertical asymptote at } x=-1$

$\textup{There are no discontinuities.}$

Correct answer:

$\textup{Hole at } x=1; \textup{ No vertical asymptotes.}$

Explanation:

Factorize the numerator for the function:

$y=\frac{x^2-1}{x-1}$

$y=\frac{(x+1)(x-1)}{x-1} = x+1$

The removable discontinuity is x-1 since this is a term that can be eliminated from the function. There are no vertical asymptotes.

Set the removable discontinutity to zero and solve for the location of the hole.

x-1=0

x=1

The hole is located at: x=1

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Example Question #2 : Find A Point Of Discontinuity

For the following function, $y=\frac{x^2-6x+9}{x^2-9}$ , find all discontinuities, if possible.

Possible Answers:

$\textup{Asymptote at: }x=\pm3$

$\textup{Hole at: }x=\pm3$

$\textup{There are no discontinuities.}$

$\textup{Asymptote at: } x=3 \textup{; Hole at: x=-3}$

$\textup{Asymptote at: } x=-3 \textup{; Hole at: x=3}$

Correct answer:

$\textup{Asymptote at: } x=-3 \textup{; Hole at: x=3}$

Explanation:

Rewrite the function $y=\frac{x^2-6x+9}{x^2-9}$ in its factored form.

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

Since the x-3 term can be cancelled, there is a removable discontinuity, or a hole, at x=3 .

The remaining denominator of x+3 indicates a vertical asymptote at x=-3 .

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Example Question #3 : Find A Point Of Discontinuity

If possible, find the type of discontinuity, if any:

$y= \frac{6x-9}{2x-3}$

Possible Answers:

$\textup{Asymptote at }x=\frac{3}{2}$

$\textup{Hole at }x=\frac{3}{2}$

$\textup{Hole and asymptote at }x=\frac{3}{2}$

$\textup{Hole at }x=-\frac{3}{2}$

$\textup{There are no discontinuities.}$

Correct answer:

$\textup{Hole at }x=\frac{3}{2}$

Explanation:

By looking at the denominator of $y= \frac{6x-9}{2x-3}$ , there will be a discontinuity.

Since the denominator cannot be zero, set the denominator not equal to zero and solve the value of .

$2x-3\neq 0$

$2x\neq 3$

$x\neq\frac{3}{2}$

There is a discontinuity at $x=\frac{3}{2}$ .

To determine what type of discontinuity, check if there is a common factor in the numerator and denominator of $y= \frac{6x-9}{2x-3}$ .

Since the common factor is existent, reduce the function.

$y= \frac{6x-9}{2x-3}=\frac{(3)(2x-3)}{2x-3}=3$

Since the 2x-3 term can be cancelled, there is a removable discontinuity, or a hole, at $x=\frac{3}{2}$ .

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Example Question #4 : Find A Point Of Discontinuity

Find the point of discontinuity for the following function:

$m(x)=\frac{x^2+4x-5}{x^2+7x+10}$

Possible Answers:

(0, 4)

(5, 1)

(-5, 2)

There is no point of discontinuity for the function.

Correct answer:

(-5, 2)

Explanation:

Start by factoring the numerator and denominator of the function.

$m(x)=\frac{x^2+4x-5}{x^2+7x+10}=\frac{(x+5)(x-1)}{(x+5)(x+2)}=\frac{x-1}{x+2}$

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

Since x=-5 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.

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

(-5, 2) 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:

$p(x)=\frac{x^3-4x^2-5x}{x^2-9x+20}$

Possible Answers:

(2, -3)

(-1, 0)

(5, 30)

There is no point fo discontinuity for this function.

Correct answer:

(5, 30)

Explanation:

Start by factoring the numerator and denominator of the function.

$p(x)=\frac{x^3-4x^2-5x}{x^2-9x+20}=\frac{x(x^2-4x-5)}{(x-5)(x-4)}=\frac{x(x-5)(x+1)}{(x-5)(x-4)}=\frac{x(x+1)}{x-4}$

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

Since x=5 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.

$\frac{5(5+1)}{5-4}=30$

(5, 30) is the point of discontinuity.

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Example Question #6 : Find A Point Of Discontinuity

Find a point of discontinuity for the following function:

$r(x)=\frac{16-4x^2}{x^2-4}$

Possible Answers:

There are no discontinuities for this function.

(2, 4)

(-2, -4)

(-4, -2)

Correct answer:

(-2, -4)

Explanation:

Start by factoring the numerator and denominator of the function.

$r(x)=\frac{16-4x^2}{x^2-4}=\frac{-4(x^2-4)}{(x+2)(x-2)}=\frac{-4(x-2)(x+2)}{(x-2)(x+2)}=-4$

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

Since $x=\pm2$ is a zero for both the numerator and denominator, there is a point of discontinuity there. Since the final function is r(x)=-4 , (2, -4) and (-2, -4) are points of discontinuity.

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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.

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

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

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

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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Precalculus : Find a Point of Discontinuity

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