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

Jun 21, 2016

for vertical asymptotes, look for when the denominator is zero. here that requires ${x}^{2} - 4 = 0 \setminus \implies x = \pm 2$
next, exploring $x \setminus \to \setminus \pm \setminus \infty$, we should note that $\frac{2 {x}^{2} + x - 3}{{x}^{2} - 4} = \frac{2 + \frac{1}{x} - 3 \setminus {x}^{2}}{1 - \frac{4}{x} ^ 2}$ , obtained by dividing top and bottom by ${x}^{2}$
and it's not hard to see that ${\left[\frac{2 + \frac{1}{x} - 3 \setminus {x}^{2}}{1 - \frac{4}{x} ^ 2}\right]}_{x \setminus \to \pm \setminus \infty} = 2$ which is a horizontal asymptote