Question

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

Answer 1

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`

`rArr x = -1 , x = 6" are the asymptotes "`

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

divide all terms on numerator/denominator by `x^2`

`((2x^2)/x^2)/(x^2/x^2 - (5x)/x^2 - 6/x^2)= 2/(1-5/x -6/x^2)`

As x → ∞ , `5/x" and " 6/x^2 → 0`

`rArr y = 2/1 = 2 " is the asymptote "`

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