Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)= (2x+4) /( x^2-3x-4)`?

Answer 1

The vertical asymptotes are `x=4` and `x=-1`. The horizontal asymptote is `y=0`. There are no oblique asymptotes.

Explanation:

`f(x)=(2x+4)/(x^2-3x-4)`

Factor the denominator.

`f(x)=frac{2x+4}{(x-4)(x+1)}`

To find the vertical asymptotes, set the denominator equal to zero and solve.
`(x-4)(x+1)=0`
`x-4=0color(white)(aaa)x+1=0`
`x=4color(white)(aaa)x=-1color(white)(aaa)`These are the VA's.

To find the horizontal asymptotes, compare the degree of the numerator to the degree of the denominator.

• If the deg of the num > deg of den, there is an oblique asymptote.

• If the deg of the num = deg of den, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.

• If the deg of the num < deg of den, the HA is `y=0`.

In this example, the degree of the numerator is `color(red)1` and the degree of the denominator is `color(blue)2`
`f(x)=(2x^color(red)1+4)/(x^color(blue)2-3x-4)`

The degree of the numerator < degree of the denominator, so the HA is `y=0`.

There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.