# How do you find the Vertical, Horizontal, and Oblique Asymptote given (8x-48)/(x^2-13x+42)?

Nov 4, 2016

An horizontal asimptote in $y = 0$ and a vertical one at $x = 7 x$

#### Explanation:

The function can be factorized at the numerator and denominator so that it becomes $f \left(x\right) = 8 \frac{x - 6}{\left(x - 6\right) \left(x - 7\right)} =$ that under the existence condition excludind $x = 6$ and $x = 7$ can be rewritten

$f \left(x\right) = \frac{8}{x - 7}$. This function has got a vertical asymptote in $x = 7$ and an horizontal one at $y = 0$ graph{8/(x-7) [-11.67, 28.33, -9.68, 10.32]}

Nov 5, 2016

vertical asymptote at x = 7
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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

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

There is an excluded value at x = 6.

solve: $x - 7 = 0 \Rightarrow x = 7 \text{ is the asymptote}$

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.

$f \left(x\right) = \frac{\frac{8}{x}}{\frac{x}{x} - \frac{7}{x}} = \frac{\frac{8}{x}}{1 - \frac{7}{x}}$

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

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

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