# How do you find the vertical, horizontal or slant asymptotes for f(x) = (4x^2 - 1 ) / (2x^2 + 5x - 12)?

Mar 16, 2016

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

#### Explanation:

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve :  2x^2 + 5x - 12 = 0 → (2x-3)(x+4) = 0

$\Rightarrow x = - 4 \text{ and " x = 3/2 " are the asymptotes }$

Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0

divide all terms on numerator/denominator by ${x}^{2}$

$\frac{\frac{4 {x}^{2}}{x} ^ 2 - \frac{1}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 + \frac{5 x}{x} ^ 2 - \frac{12}{x} ^ 2} = \frac{4 - \frac{1}{x} ^ 2}{2 + \frac{5}{x} - \frac{12}{x} ^ 2}$

now as x →∞  1/x^2 , 5/x " and " 12/x^2 → 0

rArr y = 4/2 → y = 2 " is the asymptote "

Slant asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no slant asymptotes.

Here is the graph of the function.
graph{(4x^2-1)/(2x^2+5x-12) [-10, 10, -5, 5]}