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

Feb 24, 2017

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: $\frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n}}{{b}_{m} {x}^{m}}$

If $n = m$ then the horizontal asymptote is at $y = {a}_{n} / {b}_{n}$

In the function: $f \left(x\right) = \frac{2 {x}^{3} + \ldots}{{x}^{3} + \ldots} , m = n = 3$,

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 $y = 2$