How do you find the asymptotes for $f (x) = \frac{x^{2}}{x + 5}$ $f(x)=x^2/(x+5)$ ?

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Answer 1 · 2017-08-02T15:13:38

The vertical asymptote is $x = - 5$ $x=-5$
The slant asymptote is $y = x - 5$ $y=x-5$
There is no horizontal asymptote

The vertical asymptotes are calculated by performing the limits

$lim_(x->-5^(-))f(x)=lim_(x->-5^(-))x^2/(x+5)= 25/(0^-) = -oo$

$lim_(x->-5^(+))f(x)=lim_(x->-5^(+))x^2/(x+5)= 25/(0^+) = +oo$

The vertical asymptote is $x=-5$

We perform a long division to calculate the slant asymptote

$color(white)(aaaa)$ $x+5$ $color(white)(aaaa)|$ $x^2+0x+0$ $|$ $color(white)(aaaa)$ $x-5$

$color(white)(aaaaaaaaaaaaa)$ $x^2+5x$

$color(white)(aaaaaaaaaaaaaa)$ $0-5x$

$color(white)(aaaaaaaaaaaaaaaa)$ $-5x-25$

$color(white)(aaaaaaaaaaaaaaaaaaa)$ $0+25$

Therefore,

$f(x)=x-5+25/(x+5)$

$lim_(x->-oo)f(x)-(x-5)=lim_(x->-oo)25/(x+5)=0^-$

$lim_(x->+oo)f(x)-(x-5)=lim_(x->+oo)25/(x+5)=0^+$

The slant asymptote is $y=x-5$

To determine the horizontal asymptote, we calculate

$lim_(x->-oo)f(x)=lim_(x->-oo)x^2/(x+5)=-oo$

$lim_(x->+oo)f(x)=lim_(x->+oo)x^2/(x+5)=+oo$

There is no horizontal asymptote

graph{(y-x^2/(x+5))(y-x+5)=0 [-58.5, 58.53, -29.27, 29.28]}

How do you find the asymptotes for f(x)=x^2/(x+5)f(x)=x2x+5f(x)=x^2/(x+5)?