# How do you find the vertical, horizontal and slant asymptotes of: f(x) =2/(x^2+3x-10)?

Jul 5, 2016

vertical asymptotes x = -5 , x = 2
horizontal asymptote y = 0

#### Explanation:

The denominator cannot be zero as this would yield division by zero which is undefined. Setting the denominator equal to zero and solving for x will give the values that x cannot be and if the numerator is non-zero for these x then they are vertical asymptotes.

solve: ${x}^{2} + 3 x - 10 = 0 \Rightarrow \left(x + 5\right) \left(x - 2\right) = 0$

$\Rightarrow x = - 5 , x = 2 \text{ are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest exponent of x. that is ${x}^{2}$

$\frac{\frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{3 x}{x} ^ 2 - \frac{10}{x} ^ 2} = \frac{\frac{2}{x} ^ 2}{1 + \frac{3}{x} - \frac{10}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2) Hence there are no slant asymptotes.
graph{(2)/(x^2+3x-10) [-10, 10, -5, 5]}