How do you find the vertical, horizontal and slant asymptotes of: y =(x^3 - 64)/(x^2 + x - 20)?

Nov 14, 2016

The vertical asymptote is $x = - 5$
The slant asymptote is $y = x - 1$
No horizontal asymptote

Explanation:

Let's factorise the denominator and the numerator

${x}^{2} + x - 20 = \left(x + 5\right) \left(x - 4\right)$

${x}^{3} - 64 = \left(x - 4\right) \left({x}^{2} + 4 x + 16\right)$

Therefore, $y = \frac{{x}^{3} - 64}{{x}^{2} + x - 20} = \frac{\cancel{x - 4} \left({x}^{2} + 4 x + 16\right)}{\left(x + 5\right) \cancel{x - 4}}$

$y = \frac{{x}^{2} + 4 x + 16}{x + 5}$

As we cannot divide by $0$, so $x \ne - 5$

the vertical asymptote is $x = - 5$

The degree of the numerator is $>$ the degree of the denomitor

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} + 4 x + 16$$\textcolor{w h i t e}{a a a a}$∣$x + 5$

$\textcolor{w h i t e}{a a a a}$${x}^{2} + 5 x$$\textcolor{w h i t e}{a a a a a a a a}$∣$x - 1$

$\textcolor{w h i t e}{a a a a a}$$0 - x$

$\textcolor{w h i t e}{a a a a a a a}$$- x + 16$

$\textcolor{w h i t e}{a a a a a a a}$$- x - 5$

$\textcolor{w h i t e}{a a a a a a a a a}$$0 + 21$

$\therefore y = \frac{{x}^{2} + 4 x + 16}{x + 5} = x - 1 + \frac{21}{x + 5}$

So, $y = x - 1$ is a slant asymptote

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} {x}^{2} / x = {\lim}_{x \to \pm \infty} x = \pm \infty$

So, no horizontal asymptote

graph{(y-((x^2+4x+16)/(x+5)))(y-x+1)=0 [-58.5, 58.56, -29.32, 29.23]}