How do you find the Vertical, Horizontal, and Oblique Asymptote given (2x^2-8)/(x^2+6x+8)?

Jul 16, 2016

vertical asymptote x = -4
horizontal asymptote y = 2

Explanation:

The first step is to factorise and simplify.

$\Rightarrow f \left(x\right) = \frac{2 \left(x - 2\right) \cancel{\left(x + 2\right)}}{\cancel{\left(x + 2\right)} \left(x + 4\right)} = \frac{2 x - 4}{x + 4}$

The denominator cannot be zero as this is 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 of x then it is a vertical asymptote.

solve: x + 4 = 0 → x = -4 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

$\frac{\frac{2 x}{x} - \frac{4}{x}}{\frac{x}{x} + \frac{4}{x}} = \frac{2 - \frac{4}{x}}{1 + \frac{4}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{2 - 0}{1 + 0}$

$\Rightarrow y = 2 \text{ is the asymptote}$

Oblique 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 oblique asymptotes.
graph{(2x-4)/(x+4) [-20, 20, -10, 10]}