How do you find the vertical, horizontal or slant asymptotes for $f (x) = \frac{x^{2} + x + 3}{x - 1}$ $f(x) = (x^2 + x + 3) /( x-1)$ ?

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Answer 1 · 2017-01-08T05:52:30

The vertical asymptote is $x = 1$ $x=1$
The slant asymptote is $y = x + 2$ $y=x+2$
No horizontal asymptote

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{1}$

As we cannot divide by $0$ , $x!=1$

The vertical asymptote is $x=1$

As the degree of the numerator is $>$ the degree of the denominator, we have a slant asymptote.

Let's do a long division

$color(white)(aaaa)$ $x^2+x+3$ $color(white)(aaaa)$ $∣$ $x-1$

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

$color(white)(aaaa)$ $0+2x+3$

$color(white)(aaaaaa)$ $+2x-2$

$color(white)(aaaaaaa)$ $+0+5$

Therefore,

$f(x)=(x^2+x+3)/(x-1)=(x+2)+5/(x-1)$

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

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

The slant asymptote is $y=x+2$

graph{(y-(x^2+x+3)/(x-1))(y-x-2)=0 [-32.48, 32.44, -16.2, 16.32]}

How do you find the vertical, horizontal or slant asymptotes for f(x) = (x^2 + x + 3) /( x-1)f(x)=x2+x+3x−1f(x) = (x^2 + x + 3) /( x-1)?