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

1 Answer
Nov 14, 2016

Answer:

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=(x+5)(x-4)#

#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]}