# How do you find the vertical, horizontal or slant asymptotes for (-7x + 5) / (x^2 + 8x -20)?

Feb 21, 2017

vertical asymptotes at x = -10 and x = 2
horizontal asymptote at y = 0

#### Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } {x}^{2} + 8 x - 20 = 0 \Rightarrow \left(x + 10\right) \left(x - 2\right) = 0$

$\Rightarrow x = - 10 \text{ and "x=2" are the asymptotes}$

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 power of x, that is ${x}^{2}$

$\Rightarrow \frac{\frac{- 7 x}{x} ^ 2 + \frac{5}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{8 x}{x} ^ 2 - \frac{20}{x} ^ 2} = \frac{- \frac{7}{x} + \frac{5}{x} ^ 2}{1 + \frac{8}{x} - \frac{20}{x} ^ 2}$

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

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2) Hence there are no slant asymptotes.
graph{(-7x+5)/(x^2+8x-20) [-20, 20, -10, 10]}