# How do you find the asymptotes for H(x)= (x^3-8) / (x^2-5x+6)?

Dec 27, 2016

We have one vertical asymptotes $x = 3$ and one slanting or oblique asymptote $y = x + 5$

#### Explanation:

In algebraic expressions, horizontal asymptotes are where denominator tends to infinity. Here, as denominator is quadratic, we can get up to two horizontal asymptotes.

Further, if highest degree of numerator and denominator are equal, we have vertical asymptote (which is not the case here) and if highest degree of numerator is just one greater than that of denominator, as is the case here, we get a slanting or oblique asymptote.

Now $H \left(x\right) = \frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6}$

= $\frac{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)}{\left(x - 3\right) \left(x - 2\right)} = \frac{{x}^{2} + 2 x + 4}{x - 3}$

Hence as $x \to 3$, $H \left(x\right) \to \infty$

Hence, we have horizontal asymptote $x = 3$.

Further, as $H \left(x\right) = \frac{{x}^{2} + 2 x + 4}{x - 3} = \frac{x \left(x - 3\right) + 5 \left(x - 3\right) + 19}{x - 3}$

= $x + 5 + \frac{19}{x - 3}$

= $x + 5 + \frac{\frac{19}{x}}{1 - \frac{3}{x}}$

Hence as $x \to \infty$, $H \left(x\right) \to x + 5$ and slanting / oblique asymptote is $y = x + 5$
graph{(y-(x^3-8)/(x^2-5x+6))(y-x-5)=0 [-36.7, 43.3, -10.08, 29.92]}