How do you find vertical, horizontal and oblique asymptotes for #(x^3+5x^2)/(x^2-25)#?

Question

Elijah Allen · Accepted Answer

The vertical asymptote is #x=5#
The oblique asymptote is #y=x+5#
No horizontal asymptote

The numerator is #x^2-25=(x+5)(x-5)# additionally, the numerator is #x^3+5x^2=x^2(x+5)# Consequently, #(x^3+5x^2)/(x^2-25)=(x^2cancel(x+5))/(cancel(x+5)(x-5))# #=x^2/(x-5)# Let #f(x)=x^2/(x-5)# As we cannot divide by #0#, #x!=5# The vertical asymptote is #x=5# The degree of the numerator is #># than the degree of the denominator, there is an oblique asymptote. Let's divide something long. #color(white)(aaaa)##x^2##color(white)(aaaaaaaaaaaa)##|##x-5# #color(white)(aaaa)##x^2-5x##color(white)(aaaaaaaa)##|##x+5# #color(white)(aaaaa)##0+5x# #color(white)(aaaaaaa)##+0+25# Consequently, #f(x)=x+5+25/(x-5)# #lim_(x->+oo)(f(x)-(x+5))=lim_(x->+oo)25/(x-5)=0^+# #lim_(x->-oo)(f(x)-(x+5))=lim_(x->-oo)25/(x-5)=0^-# The oblique asymptote is #y=x+5# graph{(y-x-5)(y+50x-250)=0 [-65.86, 65.94, -32.9, 32.9]}

James Allen · Answer

undefined To find the vertical asymptotes of the function $ \frac{x^3 + 5x^2}{x^2 - 25} $, identify where the denominator is equal to zero, excluding any values where the numerator is also zero.

For the denominator $x^2 - 25$, we have $x^2 - 25 = 0$. This yields $x = \pm 5$.

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

To find horizontal asymptotes, compare the degrees of the numerator and denominator polynomials:

Degree of numerator = 3
Degree of denominator = 2

Since the degree of the numerator is greater than the degree of the denominator by 1, there is no horizontal asymptote.

To find oblique asymptotes, perform polynomial division (long division or synthetic division) of the numerator by the denominator. If the result is a polynomial with a nonzero remainder, there is an oblique asymptote. If the remainder is zero, there is no oblique asymptote.

Performing polynomial division, you'll find:

$ \frac{x^3 + 5x^2}{x^2 - 25} = x + 5 + \frac{125x}{x^2 - 25} $

Since there's a nonzero remainder (125x), there is an oblique asymptote.

Therefore, the vertical asymptotes are $x = -5$ and $x = 5$, there is no horizontal asymptote, and there is an oblique asymptote.