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

Aug 5, 2016

vertical asymptotes $x = - 4 , x = - \frac{1}{2}$
horizontal asymptote $y = \frac{3}{2}$

#### Explanation:

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are asymptotes.

solve: 2x^2+9x+4)=0rArr(2x+1)(x+4)=0

$\Rightarrow x = - 4 \text{ and " x=-1/2" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$\frac{\frac{3 {x}^{2}}{x} ^ 2 - \frac{3 x}{x} ^ 2 - \frac{36}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 + \frac{9 x}{x} ^ 2 + \frac{4}{x} ^ 2} = \frac{3 - \frac{3}{x} - \frac{36}{x} ^ 2}{2 + \frac{9}{x} + \frac{4}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{3 - 0 - 0}{2 + 0 + 0}$

$\Rightarrow y = \frac{3}{2} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no slant asymptotes.
graph{(3x^2-3x-36)/(2x^2+9x+4) [-20, 20, -10, 10]}