How do you find the Vertical, Horizontal, and Oblique Asymptote given H(x)= (x^3-8) / (x^2-5x+6)?

Aug 4, 2016

Vertical: x=3; no Horizontal asymptote; Oblique: y=x+5

Explanation:

1) The vertical asymptotes depend on the domain; the domain is obtained by solving the following:

${x}^{2} - 5 x + 6 \ne 0$

that is solved by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$

where a=1; b=-5; c=6

then

$x = \frac{5 \pm \sqrt{{5}^{2} - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

$= \frac{5 \pm \sqrt{25 - 24}}{2}$

$= \frac{5 \pm 1}{2}$

x_1=2;x_2=3

The domain of the given function is:

$x \ne 2 \mathmr{and} x \ne 3$

Now let's calculate

${\lim}_{x \to 2} \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} = {\lim}_{x \to 2} \frac{\left(\cancel{x - 2}\right) \left({x}^{2} + 2 x + 4\right)}{\left(\cancel{x - 2}\right) \left(x - 3\right)} = - 12$

and

${\lim}_{x \to 3} \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} = {\lim}_{x \to 3} \frac{\left(\cancel{x - 2}\right) \left({x}^{2} + 2 x + 4\right)}{\left(\cancel{x - 2}\right) \left(x - 3\right)} = \infty$

Then the vertical asymptote is the line x=3

2) Let's calculate

${\lim}_{x \to \infty} \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} = \infty$

then there are no horizontal asymptote

3) Let's calculate

$m = {\lim}_{x \to \infty} \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} \cdot \frac{1}{x} = \frac{{x}^{3} - 8}{{x}^{3} - 5 {x}^{2} + 6 x} = 1$

that's the slope of the oblique asymptote.

Let's calculate the intercept

$n = {\lim}_{x \to \infty} \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} - m x = {\lim}_{x \to \infty} \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} - x$

$n = {\lim}_{x \to \infty} \frac{{\cancel{x}}^{3} - 8 - {\cancel{x}}^{3} + 5 {x}^{2} - 6 x}{{x}^{2} - 5 x + 6} = 5$

Then the oblique asymptote is the line

$y = x + 5$

graph{(x^3-8)/(x^2-5x+6) [-20, 10, -15, 5]}