Find Intercepts and Asymptotes - Pre-Calculus

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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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Question

Find the vertical asymptotes of the following rational function.

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

Answer

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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Question

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

Answer

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:

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

Since it's not possible to substitute 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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Question

What is the -intercept of the following function?

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

Answer

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

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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Question

What is a vertical asymptote of the following function?

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

Answer

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

With our function, we complete this process.

The denominator is , so we begin:

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Question

Find the zeros and asymptotes for

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

Answer

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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Question

Find the slant and vertical asymptotes for the equation

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

Answer

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

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

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

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

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

The slant asymptote is

.

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Question

Find the slant asymptote for

$y = \frac{2x^2 - 13 x - 7 }{ x - 7 }$ .

Answer

By factoring the numerator, we see that this equation is equivalent to

$y = \frac{(2x+1)(x - 7 )}{x - 7}$ .

That means that we can simplify this equation to .

That means that isn't the slant asymptote, but the equation itself.

is definitely an asymptote, but a vertical asymptote, not a slant asymptote.

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Question

Which of these functions has a vertical asymptote of and a slant asymptote of ?

Answer

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

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

Dividing the first one gives us with no remainder.

Dividing the last one gives us with a remainder.

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

The answer is

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

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Question

Find the y-intercept of $y=\frac{3-x}{x-3}+3$ , if any.

Answer

Be careful not to confuse this equation with the linear slope-intercept form. The y-intercept of an equation is the y-value when the x-value is zero.

Substitute the value of into the equation.

$y=\frac{3-0}{0-3}+3$

Simplify the equation.

$y=\frac{3-0}{0-3}+3 = \frac{3}{-3}+3 = -1+3 = 2$

The y-intercept is:

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Question

Find the horizontal asymptote of the function:

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

Answer

To find the horizontal asymptote, take the leading term of the numerator and the denominator and divide. In this case:

$f(x) = \frac{x^2}{x^2}$

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Question

Find the vertical and horizontal asymptotes of the function

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

Answer

The function

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

is already in simplified form.

To find the vertical asymptotes, we set the denominator equal to and solve for .

yields the vertical asymptotes

To find the horizontal asymptote, we examine the largest degree of between the numerator and denominator

Note that

Because the largest degree of in the numerator is less than the largest degree of in the denominator, or

we find the horizontal symptote to be

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Question

Determine the asymptotes, if any: $y=\frac{x^2-2x+1}{x^2-1}$

Answer

Factorize both the numerator and denominator.

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

Notice that one of the binomials will cancel.

The domain of this equation cannot include $x=\pm1$ .

The simplified equation is:

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

Since the term canceled, the term will have a hole instead of an asymptote.

Set the denominator equal to zero.

Subtract one from both sides.

There will be an asymptote at only:

The answer is:

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Question

Where is an asymptote located, if any? $y=\frac{2x-4}{x^2-4}$

Answer

Factor the numerator and denominator.

Rewrite the equation.

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

Notice that the will cancel. This means that the root of will be a hole instead of an asymptote.

Set the denominator equal to zero and solve for x.

An asymptote is located at:

The answer is:

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Question

Find the vertical and horizontal asymptotes of the function

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

Answer

The function

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

is already in simplified form.

To find the vertical asymptotes, we set the denominator equal to and solve for .

yields the vertical asymptotes

To find the horizontal asymptote, we examine the largest degree of between the numerator and denominator

Note that

Because the largest degree of in the numerator is less than the largest degree of in the denominator, or

we find the horizontal symptote to be

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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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