How do you find the vertical, horizontal or slant asymptotes for f(x)= (x-5)/(x^2+6x+5)?

Jun 6, 2016

vertical asymptotes x = -5 , x = -1
horizontal asymptote y = 0

Explanation:

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

solve: ${x}^{2} + 6 x + 5 = 0 \Rightarrow \left(x + 5\right) \left(x + 1\right) = 0$

$\Rightarrow x = - 5 , x = - 1 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by ${x}^{2}$

(x/x^2-5/x^2)/(x^2/x^2+(6x)/x^2+5/x^2)=(1/x-5/x^2)/(1+6/x+5/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 1 , denominator-degree 2 ). Hence there are no slant asymptotes.
graph{(x-5)/(x^2+6x+5) [-8.89, 8.89, -4.444, 4.445]}