# How do you find the vertical, horizontal or slant asymptotes for f(x)= (-10x+3)/(8x+2)?

Jun 2, 2016

vertical asymptote $x = - \frac{1}{4}$
horizontal asymptote $y = - \frac{5}{4}$

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : 8x + 2 = 0 → 8x = -2 $\Rightarrow x = \frac{- 2}{8} = - \frac{1}{4}$

$\Rightarrow x = - \frac{1}{4} \text{ is the asymptote}$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$

divide terms on numerator/denominator by x

$\frac{\frac{- 10 x}{x} + \frac{3}{x}}{\frac{8 x}{x} + \frac{2}{x}} = \frac{- 10 + \frac{3}{x}}{8 + \frac{2}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{- 10 + 0}{8 + 0}$

$\Rightarrow y = - \frac{10}{8} = - \frac{5}{4} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1). Hence there are no slant asymptotes.
graph{(-10x+3)/(8x+2) [-10, 10, -5, 5]}