How do you find the vertical, horizontal or slant asymptotes for $f(x) = (-x^2 + 4x)/(x+2)$ ?

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Answer 1 · 2017-01-06T11:00:02

The vertical asymptote is $x=-2$
The slant asymptote is $y=-x+6$
No horizontal asymptote

The domain of $f(x)$ is $D_f(x)=RR-{-2}$

As we cannot divide by $0$ , $x!=-2$

The vertical asymptote is $x=-2$

The degree of the numerator is $>$ than the degree of the denominator, so there is a slant asymptote.

To find the slant, we start by doing a long division

$color(white)(aaaa)$ $-x^2+4x$ $color(white)(aaaa)$ $∣$ $x+2$

$color(white)(aaaa)$ $-x^2-2x$ $color(white)(aaaa)$ $∣$ $-x+6$

$color(white)(aaaaaaa)$ $0+6x$

$color(white)(aaaaaaaaa)$ $+6x+12$

$color(white)(aaaaaaaaaaaaaa)$ $-12$

So,

$f(x)=(-4x^2+4x)/(x+2)=-x+6-12/(x+2)$

$lim_(x->-oo)f(x)+(x-6)=lim_(x->-oo)-12/(x+2)=0^+$

$lim_(x->+oo)f(x)+(x-6)=lim_(x->+oo)-12/(x+2)=0^-$

The slant asymptote is $y=-x+6$

No horizontal asymptote

graph{(y-(-x^2+4x)/(x+2))(y+x-6)(y-25x-50)=0 [-35, 38.04, -14.36, 22.2]}

How do you find the vertical, horizontal or slant asymptotes for f(x) = (-x^2 + 4x)/(x+2)f(x) = (-x^2 + 4x)/(x+2)?