Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)= -1/(x+1)^2`?

Answer 1

vertical asymptote: `x=-1`
horizontal asymptote: `f(x)=0`
oblique asymptote: does not exist

Explanation:

Finding the Vertical Asymptote

Given,

`f(x)=-1/(x+1)^2`

Set the denominator equal to `0` and solve for `x`.

`(x+1)^2=0`

`x+1=0`

`color(green)(|bar(ul(color(white)(a/a)color(black)(x=-1)color(white)(a/a)|)))`

Finding the Horizontal Asymptote

Given,

`f(x)=-1/(x+1)^2`

If you foil the denominator, you will notice that the degree of the denominator is `2`.

`f(x)=-1/(x^2+2x+1)`

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is `f(x)=0`.

`color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=0)color(white)(a/a)|)))`

Finding the Oblique Asymptote

Given,

`f(x)=-1/(x+1)^2`

There would be a slant asymptote if the degree of the leading term in the numerator is `1` value larger than the degree of the leading term in the denominator. In your case, we see the opposite — ie. the degree in the denominator is greater than the degree in the numerator.

`:.`, the oblique asymptote does not exist.