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

May 7, 2016

vertical asymptotes x = ± 2
horizontal asymptote y = 3

#### Explanation:

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

solve : x^2-4=0 → (x-2)(x+2)=0 → x=±2

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

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$

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

$\frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{3 + \frac{2}{x} - \frac{1}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

as $x \to \pm \infty , y \to \frac{3 + 0 - 0}{1 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 2) hence there are no slant asymptotes.
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}