Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (3x^2 + 4)/(x+1)`?

Answer 1

Vertical asymptote at `x= -1`, Slant asymptote: `y=3x-3` ,
Horizontal asymptote: Absent

Explanation:

`f(x) =(3x^2+4)/(x+1)`
Vertical asymptote : Denominator is `0 :. x+1 = 0 or x = -1`

Vertical asymptote at `x= -1`

Degree i.e maximum power of `x` in numerator is `1` more than denominator, so there is no horizontal asymptote.

Slant asymptote: Degree i.e maximum power of `x` in numerator is `1` more than denominator , so there is slant asymptote.

By long division of (f(x) we get dividend as `y=3x-3` and remainder `7 :. y = 3x-3` is the slant asymptote.

graph{(3x^2+4)/(x+1) [-80, 80, -40, 40]} [Ans]