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Precalculus : Find Intercepts and Asymptotes

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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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Example Question #2 : Find Intercepts And Asymptotes

Find the vertical asymptotes of the following rational function.

$\small \frac{8x^2}{4x^2+31x-8}$

Possible Answers:

$\small x=4,\frac{1}{8}$

$\small x=-\frac{1}{8},4$

$\small x=\frac{1}{4}, -8$

$\small \small x=-\frac{1}{4},8$

No vertical asymptotes.

Correct answer:

$\small x=\frac{1}{4}, -8$

Explanation:

Finding the vertical asymptotes of the rational function $\small \frac{8x^2}{4x^2+31x-8}$ amounts to finding the roots of the denominator, $\small 4x^2+31x-8$ .

It is easy to check, using the quadratic formula,

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

$x=\frac{-31\pm\sqrt {31^2-4(4)(-8)}}{2(4)}$

$x=\frac{-31\pm\sqrt {961+128}}{8}$

$x=\frac{-31\pm 33}{8}=\frac{1}{4}, -8$

that the roots, and thus the asymptotes, are $\small x=\frac{1}{4}, -8$ .

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Example Question #3 : Find Intercepts And Asymptotes

Find the y-intercept and asymptote, respectively, of the following function, if possible. $y=\frac{3x}{x}$

Possible Answers:

$\textup{The y-intercept does not exist. Asymptote at: }x=0$

$\textup{There are no y-intercepts or asymptotes.}$

$\textup{y-intercept at: }(0,3);\:\textup{Asymptote at: }y=0$

$\textup{y-intercept at: }(0,3);\:\textup{Asymptote at: }y=3$

$\textup{y-intercept at: }(0,3);\:\textup{Asymptote at: }x=0$

Correct answer:

$\textup{There are no y-intercepts or asymptotes.}$

Explanation:

Before we start to simplify the problem, it is crucial to immediately identify the domain of this function $y=\frac{3x}{x}$ .

The denominator cannot be zero, since it is undefined to divide numbers by this value. After simplification, the equation is:

y=3

The domain is $x\neq0$ and there is a hole at x=0 since there is a removable discontinuity. There are no asymptotes.

Since it's not possible to substitute x=0 into the original equation, the y-intercept also does not exist.

Therefore, the correct answer is:

$\textup{There are no y-intercepts or asymptotes.}$

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Example Question #1 : Find Intercepts And Asymptotes

What is a vertical asymptote of the following function?

$\frac{x^2 + 3x + 6}{x + 2}$

Possible Answers:

y = 2

x = -2

x = -3

y = 3

x = 3

Correct answer:

x = -2

Explanation:

To find the vertical asymptote of a function, we set the denominator equal to .

With our function, we complete this process.

The denominator is x + 2 , so we begin:

x + 2 = 0

x = -2

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Example Question #7 : Find Intercepts And Asymptotes

What is the -intercept of the following function?

$\frac{x^2 + 3x + 6}{x + 2}$

Possible Answers:

There is no -intercept.

Correct answer:

Explanation:

The y-intercept of a function is always found by substituting in x = 0 .

We can go through this process for our function.

$\frac{x^2 + 3x + 6}{x + 2}= \frac{0^2 + 3\cdot 0+ 6}{0 + 2} = \frac{ 6}{ 2} = 3$

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Example Question #8 : Find Intercepts And Asymptotes

Which of these functions has a vertical asymptote of x = 3 and a slant asymptote of y = x - 8 ?

Possible Answers:

$y = \frac{x^2 - 11x + 24}{x - 3 }$

$y = \frac{x^2 - 11x + 7 }{ x - 3 }$

$y = \frac{x^2 - 11x + 9 }{x + 3 }$

$y = \frac{x^2 + 11x + 9 }{x - 3 }$

$y = \frac{x^2 + 11x - 7 }{x + 3 }$

Correct answer:

$y = \frac{x^2 - 11x + 7 }{ x - 3 }$

Explanation:

In order for the vertical asymptote to be x = 3 , we need the denominator to be x -3 . This gives us three choices of numerators:

x^2 - 11x + 24

x^2 - 11x + 7

x^2 + 11x + 9

If the slant asymptote is x - 8 , we will be able to divide our numerator by x - 3 and get x - 8 with a remainder.

Dividing the first one gives us x - 8 with no remainder.

Dividing the last one gives us x + 14 with a remainder.

The middle numerator would give us what we were after, x - 8 with a remainder of -17.

The answer is

$y = \frac{x^2 - 11x + 7 }{ x - 3 }$

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Example Question #1 : Find Intercepts And Asymptotes

Find the zeros and asymptotes for

$y = \frac{2x^2 + 5 x - 3 }{2x^2 + 3x - 2 }$ .

Possible Answers:

Zero: x = -3 ; Asymptote: x = -2

Zeros: x = 3, -1 ; Asymptote: x = 2

Zero: $x = \frac{1}{2}$ ; Asymptotes: $x = 2, -\frac{1}{2}$

Zero: x = -3 ; Asymptotes: $x = -2, \frac{1}{2}$

Zeros: $x = 2, \frac{1}{2}$ ; Asymptotes: $x = 3, \frac{1}{2}$

Correct answer:

Zero: x = -3 ; Asymptote: x = -2

Explanation:

To find the information we're looking for, we should factor this equation:

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

This means that it simplifies to $y = \frac{x + 3 }{ x + 2 }$ .

When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.

$\\x+3=0\\x=-3$

To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.

$\\x+2=0\\x=-2$

Therefore our equation has a zero at -3 and an asymptote at -2.

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Example Question #10 : Find Intercepts And Asymptotes

Find the slant and vertical asymptotes for the equation

$y = \frac{2x^2 - 5 x + 1 }{x + 3 }$ .

Possible Answers:

Vertical asymptote: x = -3 ; Slant asymptote: y = 2 x + 1

Vertical asymptote: x = -3 ; Slant asymptote: y = 2x^2 - 11 x + 34

Vertical asymptote: x = 3 ; Slant asymptote: y = 2x + 1

Vertical asymptote: x = -3 ; Slant asymptote: y = 2x -1

Vertical asymptote: x = -3 ; Slant asymptote: y = 2x - 8

Correct answer:

Vertical asymptote: x = -3 ; Slant asymptote: y = 2x - 8

Explanation:

To find the vertical asymptote, just set the denominator equal to 0:

x + 3 = 0

x = -3

To find the slant asymptote, divide the numerator by the denominator, but ignore any remainder. You can use long division or synthetic division.

2x-8

$\\x+3\ |\overline{2 x^ 2 - 5 x + 1 }$

$\\ \underline{-(2x^2+3x) }$

-8x+1

$\underline{-(-8x-24)}$

The slant asymptote is

y = 2x - 8 .

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Precalculus : Find Intercepts and Asymptotes

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find Intercepts And Asymptotes

Example Question #1 : Find Intercepts And Asymptotes

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