Question

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

Answer 1

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`