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

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Answer 1 · 2017-02-24T23:11:44

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

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 $y = 2$

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