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

May 8, 2016

Vertical asymptotes are $x = - \frac{1}{2}$ and $x = - 4$

and horizontal asymptote $y = \frac{3}{2}$

#### Explanation:

In $\frac{3 {x}^{2} - 3 x - 36}{2 {x}^{2} + 9 x + 4}$, we get vertical asymptotes by putting denominator as zero i.e. they are given by

$2 {x}^{2} + 9 x + 4 = 0$ and factorizing by splitting

$2 {x}^{2} + 8 x + x + 4 = 0$ or $2 x \left(x + 4\right) + 1 \left(x + 4\right) = 0$ or $\left(2 x + 1\right) \left(x + 4\right) = 0$

Hence vertical asymptotes are $x = - \frac{1}{2}$ and $x = - 4$.

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

$y = \frac{3 {x}^{2}}{2 {x}^{2}} = \frac{3}{2}$

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]}