# How do you find the asymptotes for Q(x) = (2x^2)/ (x^2 - 5x - 6)?

Mar 26, 2016

vertical asymptotes x = -1 , x = 6
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 : x^2 - 5x - 6 = 0 → (x-6)(x+1) = 0

$\Rightarrow x = - 1 , x = 6 \text{ are the asymptotes }$

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

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

$\frac{\frac{2 {x}^{2}}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{5 x}{x} ^ 2 - \frac{6}{x} ^ 2} = \frac{2}{1 - \frac{5}{x} - \frac{6}{x} ^ 2}$

As x → ∞ ,  5/x" and " 6/x^2 → 0

$\Rightarrow y = \frac{2}{1} = 2 \text{ is the asymptote }$

Here is the graph of the function.
graph{(2x^2)/(x^2-5x-6) [-20, 20, -10, 10]}