# How do you find the vertical, horizontal and slant asymptotes of: y=(8x-48)/(x^2-13x+42)?

Jul 1, 2016

vertical asymptote x = 7
horizontal asymptote y = 0

#### Explanation:

The first step here is to factorise and simplify y.

$\frac{8 x - 48}{{x}^{2} - 13 x + 42} = \frac{8 \cancel{\left(x - 6\right)}}{\left(x - 7\right) \cancel{\left(x - 6\right)}} = \frac{8}{x - 7}$

The denominator of this rational function cannot be zero as this would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the value that x cannot be and if the numerator is also non-zero for this value of x then it must be a vertical asymptote.

solve : x - 7 = 0 → x = 7 is the asymptote

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$\frac{\frac{8}{x}}{\frac{x}{x} - \frac{7}{x}} = \frac{\frac{8}{x}}{1 - \frac{7}{x}}$

as $x \to \pm \infty . y \to \frac{0}{1 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0 ,denominator-degree 1 )Hence there are no slant asymptotes.
graph{8/(x-7) [-20, 20, -10, 10]}