Question

How do you find vertical, horizontal and oblique asymptotes for `f(x) = (3x^2 - 3x- 36) / (2x^2 + 9x +4)`?

Answer 1

Vertical asymptotes are `x=-1/2` and `x=-4`

and horizontal asymptote `y=3/2`

Explanation:

In `(3x^2-3x-36)/(2x^2+9x+4)`, we get vertical asymptotes by putting denominator as zero i.e. they are given by

`2x^2+9x+4=0` and factorizing by splitting

`2x^2+8x+x+4=0` or `2x(x+4)+1(x+4)=0` or `(2x+1)(x+4)=0`

Hence vertical asymptotes are `x=-1/2` and `x=-4`.

As degree of numerator is equal to that of denominator in the algebraic expression, we have only horizontal asymptote given by

`y=(3x^2)/(2x^2)=3/2`

Had the degree of numerator just one greater than that of denominator, we would have oblique asymptote.

graph{(3x^2-3x-36)/(2x^2+9x+4) [-20, 20, -10, 10]}