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

Question

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

Serenity Johnson · Accepted Answer

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

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]}

Chloe Alexander · Answer

undefined To find the vertical asymptotes of the function $ y = \frac{x^3 - 64}{x^2 + x - 20} $, we need to look for values of $ x $ that make the denominator zero but not the numerator. By factoring the denominator, we can find these values.

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

Setting each factor equal to zero gives us:

$$ x - 4 = 0 \Rightarrow x = 4 $$
$$ x + 5 = 0 \Rightarrow x = -5 $$

So, there are vertical asymptotes at $ x = 4 $ and $ x = -5 $.

To find horizontal asymptotes, we look at the degree of the numerator and the denominator of the rational function. Since the degree of the numerator is 3 and the degree of the denominator is 2, there is no horizontal asymptote.

For slant asymptotes, if the degree of the numerator is one more than the degree of the denominator, there is a slant asymptote. To find it, we perform polynomial long division or use synthetic division to divide the numerator by the denominator.

$$ \frac{x^3 - 64}{x^2 + x - 20} = x - 9 + \frac{45x - 236}{x^2 + x - 20} $$

As $ x $ approaches infinity, $ \frac{45x - 236}{x^2 + x - 20} $ approaches zero, so the slant asymptote is $ y = x - 9 $.