Rational Functions - Pre-Calculus

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Question

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

Answer

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 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.

The hole is located at:

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Question

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

Answer

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 term can be cancelled, there is a removable discontinuity, or a hole, at .

The remaining denominator of indicates a vertical asymptote at .

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Question

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

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

Answer

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 .

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

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

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Question

Find a point of discontinuity in the following function:

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

Answer

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 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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Question

Find the point of discontinuity for the following function:

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

Answer

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 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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Question

Find the point of discontinuity for the following function:

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

Answer

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 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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Question

Find the point of discontinuity for the following function:

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

Answer

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 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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Question

Find the point of discontinuity for the following function:

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

Answer

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 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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Question

Find the point of discontinuity for the following function:

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

Answer

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 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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Question

Find the point of discontinuity for the following function:

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

Answer

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 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.

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

is the point of discontinuity.

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Question

Find the point of discontinuity for the following function:

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

Answer

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 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.

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

is the point of discontinuity.

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Question

Find the point of discontinuity for the following function:

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

Answer

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 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.

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

is the point of discontinuity.

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Question

Find the point of discontinuity for the following function:

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

Answer

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 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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Question

Find a point of discontinuity for the following function:

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

Answer

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 , and are points of discontinuity.

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Question

Find a point of discontinuity for the following function:

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

Answer

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 , and are points of discontinuity.

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Question

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

Answer

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

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

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 can be cancelled, there will not be any vertical asymptotes. Even though the rational function simplifies to , 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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Question

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

Answer

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

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Question

Suppose the function below has an oblique (i.e. slant asymptote) at .

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

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

Answer

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 .

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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Question

Find the -intercept and asymptote, if possible.

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

Answer

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

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

The y-intercept is 1.

The numerator, , can be simplified by factoring it into two binomials.

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

There is a removable discontinuity at , but there are no asymptotes at since the terms can be canceled.

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

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Question

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

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

Answer

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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