# How do you find vertical, horizontal and oblique asymptotes for (x^2-2x-3) /( 2x^2-x-10)?

Jul 1, 2016

vertical asymptotes at $x = - 2 , 2.5$

horizontal asymptote is $y = \frac{1}{2}$

#### Explanation:

$\frac{{x}^{2} - 2 x - 3}{2 {x}^{2} - x - 10}$

looking at denominator

$2 {x}^{2} - x - 10 = 2 \left\{{x}^{2} - \frac{1}{2} x - 5\right\}$

$= 2 \left\{\left(x + 2\right) \left(x - \frac{5}{2}\right)\right\}$ so the denominator is zero at $x = - 2 , 2.5$

the numerator factorises as ${x}^{2} - 2 x - 3 = \left(x - 3\right) \left(x + 1\right)$ so it is zero at x = -1, 3 thus finite at $x = - 2 , 2.5$

we can therefore conclude vertical asymptotes at $x = - 2 , 2.5$

${\lim}_{x \to \pm \infty} \frac{{x}^{2} - 2 x - 3}{2 {x}^{2} - x - 10}$

$= {\lim}_{x \to \pm \infty} \frac{1 - \frac{2}{x} - \frac{3}{x} ^ 2}{2 - \frac{1}{x} - \frac{10}{x} ^ 2} = \frac{1}{2}$

so the horizontal asymptote is $y = \frac{1}{2}$