How do you find the vertical, horizontal and slant asymptotes of: $f (x) = \frac{x^{2} + 6 x - 9}{x - 2}$ $f(x) =(x^2+6x-9)/(x-2)$ ?

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Answer 1 · 2016-07-03T06:42:16

V.A.: x=2, S.A.: y=x+8

The vertical asymptote depends on the domain of the function, then since it must be:

$x \neq 2$ $x!=2$ ,

the vertical asymptote is the line $x = 2$ $x=2$

graph{(x^2+6x-9)/(x-2) [-10, 10, -5, 5]}

In fact

$lim_(x rarr2)(x^2+6x-9)/(x-2)=7/0=oo$

There are no horizontal asymptote, because

$lim_(x rarroo)(x^2+6x-9)/(x-2)=oo$

Maybe the function has a slant asymptote, so let's calculate

$m=lim_(x rarroo)(x^2+6x-9)/(x-2)*1/x=1$

and

$n=lim_(x rarr2)(x^2+6x-9)/(x-2)-mx=$

$=lim_(x rarr2)(x^2+6x-9)/(x-2)-x=$

$=lim_(x rarr2)(cancelx^2+6x-9-cancelx^2+2x)/(x-2)=$

$=lim_(x rarr2)(8x-9)/(x-2)=8$

So the equation of the slant asymptote is

$y=x+8$

How do you find the vertical, horizontal and slant asymptotes of: f(x) =(x^2+6x-9)/(x-2)f(x)=x2+6x−9x−2 f(x) =(x^2+6x-9)/(x-2)?