How do you find the vertical, horizontal and slant asymptotes of: # f(x)=((x-2)^2(2x+5)) /( (x-3)^2(x-2))#?

1 Answer
Feb 24, 2017

Answer:

Vertical asymptote at #x = 3#
Hole at #x = 2#
No slant asymptote
Horizontal asymptote at #y = 2#

Explanation:

Holes come when you can cancel a term because it is in the numerator and denominator: #x - 2 = 0; x = 2#

Vertical asymptotes come from setting all terms except the term that causes the hole #D(x) = 0 : x-3 = 0; x = 3#

Horizontal asymptotes are determined by the degrees of the numerator and denominator: #(N(x))/(D(x)) = (a_n x^n)/(b_m x^m)#

If #n = m# then the horizontal asymptote is at #y = a_n/b_n#

In the function: #f(x) = (2x^3 + ...)/(x^3 + ...), m = n = 3#,

so there is a horizontal asymptote at #y = 2#

To have a slant asymptote : #m+1 = n#

To summarize:
Vertical asymptote at #x = 3#
Hole at #x = 2#
No slant asymptote
Horizontal asymptote at #y = 2#