How do you graph y=(2x^2-9x-5)/(x^2-16) using asymptotes, intercepts, end behavior?

Oct 20, 2016

The vertical asymptotes are $x = 4$ and $x = - 4$
And the horizontal asymptote is $y = 2$

Explanation:

Factorising numerator and denominator
$y = \frac{\left(2 x + 1\right) \left(x - 5\right)}{\left(x + 4\right) \left(x - 4\right)}$
To factorise you can use the formula$x = \frac{- b \pm \left({b}^{2} - 4 a c\right)}{2 a}$
We can see that $x \ne \pm 4$ as we cannot divide by 0. Hence we obtain the vertical asymptotes $x = 4$ and $x = - 4$
And the limit of y as $x \to \infty$ is 2. therefore the horizontal asymptote is $y = 2$
When $x = 0$ , $y = \frac{5}{16}$
graph{(2x^2-9x-5)/(x^2-16) [-20, 20, -10, 10]}