Pre-Calculus - Find Intercepts and Asymptotes

Find the zeros and asymptotes for

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

Zero: ; Asymptote:

Zeros: ; Asymptote:

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

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

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

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.

Find the slant asymptote for

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

This graph does not have a slant asymptote.

Explanation

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.

Find the slant and vertical asymptotes for the equation

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

Vertical asymptote: ; Slant asymptote:

Explanation

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

What is the -intercept of the following function?

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

There is no -intercept.

Explanation

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

We can go through this process for our function.

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

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

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

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

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

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

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

Explanation

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

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

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 ?

$n = m+1, \textup{and} , \frac{a}{b}=3$

$n = m-1, \textup{and} , \frac{a}{b}=3$

$n = m+1, \textup{and} , \frac{b}{a}=3$

$n = m, \textup{and} , \frac{b}{a}=3$

$n = m-1, \textup{and} , \frac{b}{a}=3$

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 .

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.

Find the -intercept and asymptote, if possible.

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

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

$\textup{Y-intercept at: } y=3; \textup{ No asymptotes.}$

$\textup{Y-intercept at: } y=1; \textup{ Asymptote at: }x=3$

$\textup{Y-intercept at: } y=1; \textup{ Asymptote at: }x=-3$

$\textup{Y-intercept at: } y=3; \textup{ Asymptote at: }x=-3$

Explanation

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

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

$\infty$

$\textup{The y-intercept does not exist.}$

Explanation

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:

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

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

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

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

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

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

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:

The domain is 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:

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

Find the horizontal asymptote of the function:

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

$y = \frac{1}{2}$

Explanation

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

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

Find Intercepts and Asymptotes

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