# Consider f(x) = \frac{2x^2 - 8}{x^2 - 5x - 6} \\ (a) Find the domain for this function.\\ (b)...

## Question:

Consider {eq}f(x) = \frac{2x^2 - 8}{x^2 - 5x - 6}{/eq}

(a) Find the domain for this function.

(b) Find the x - and y-intercepts.

(c) Find the horizontal and vertical asymptotes.

## Asymptotes:

If {eq}\displaystyle \lim_{x\rightarrow\infty}f(x)=L \text{ OR } \lim_{x\rightarrow -\infty}f(x)=L, {/eq} then f has a horizontal asymptote at y=L.

With rational functions, we can follow these rules:

(1) If the degree of the function in the numerator is bigger than the degree of the function in the denominator, then the function will not have any horizontal asymptotes. (This is because the leading term in the numerator will dominate as x gets really large, and thus the limit as described above will be positive or negative infinity.)

(2) if the degree of the function in the denominator is bigger than the degree of the function in the numerator, then the function will have a horizontal asymptote at y=0. (This is because the leading term in the denominator will dominate as x gets really large, and thus the limit described will be zero.

(3) If the degree of the function in the numerator is equal to the degree of the function in the denominator, then you divide the leading coefficient of the top by the leading coefficient of the bottom to get a, and y=a is the horizontal asymptote. (This is because the leading terms of the numerator and the denominator will dominate the function, and the {eq}x^n {/eq} will cancel out and leave you with just the coefficients.

Any value a for which {eq}\lim_{x\rightarrow a^+}f(x)= \pm \infty \text{ or } \lim_{x\rightarrow a^-}f(x)= \pm \infty {/eq} is a vertical asymptote of the function.

Rational functions are a special case of this rule: Suppose {eq}f(x)=\frac{p(x)}{q(x)}. {/eq}

If {eq}q(a)=0, \text{ but } p(a) \neq 0, {/eq} then we say that x=a is a vertical asymptote for the function.

First, we will factor this function. This will help us in many pieces of this problem.

{eq}\displaystyle...

Become a Study.com member to unlock this answer! Create your account 