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

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Answer 1 · 2016-11-14T13:30:18

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

Let's factorise the denominator and the numerator

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

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

Therefore, $y=(x^3-64)/(x^2+x-20)=(cancel(x-4)(x^2+4x+16))/((x+5)cancel(x-4))$

$y=(x^2+4x+16)/(x+5)$

As we cannot divide by $0$ , so $x!=-5$

the vertical asymptote is $x=-5$

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

Let's do a long division

$color(white)(aaaa)$ $x^2+4x+16$ $color(white)(aaaa)$ $∣$ $x+5$

$color(white)(aaaa)$ $x^2+5x$ $color(white)(aaaaaaaa)$ $∣$ $x-1$

$color(white)(aaaaa)$ $0-x$

$color(white)(aaaaaaa)$ $-x+16$

$color(white)(aaaaaaa)$ $-x-5$

$color(white)(aaaaaaaaa)$ $0+21$

$:. y=(x^2+4x+16)/(x+5)=x-1+21/(x+5)$

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

$lim_(x->+-oo)y=lim_(x->+-oo)x^2/x=lim_(x->+-oo)x=+-oo$

So, no horizontal asymptote

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

How do you find the vertical, horizontal and slant asymptotes of: y =(x^3 - 64)/(x^2 + x - 20)y=x3−64x2+x−20y =(x^3 - 64)/(x^2 + x - 20)?