How do you find the asymptotes for (2x^2 + x + 2) / (x + 1)2x2+x+2x+1?

1 Answer
Jan 10, 2017

The vertical asymptote is x=-1x=1
The slant asymptote is y=2x-1y=2x1
No horizontal asymptote

Explanation:

Let f(x)=(2x^2+x+2)/(x+1)f(x)=2x2+x+2x+1

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

As you cannot divide by 0, x!=-1

The vertical asymptote is x=-1

As we the degree of the numerator is > than the degree of the denominator, there is a slant asymptote.

Let's do a long division

color(white)(aaaa)2x^2+x+2color(white)(aaaa)x+1

color(white)(aaaa)2x^2+2xcolor(white)(aaaaaaa)2x-1

color(white)(aaaaa)0-x+2

color(white)(aaaaaaa)-x-1

color(white)(aaaaaaaaa)0+3

Therefore,

f(x)=(2x-1)+(3)/(x+1)

So,

lim_(x->-oo)(f(x)-(2x-1))=lim_(x->-oo)3/(x+1)=0^-

lim_(x->+oo)(f(x)-(2x-1))=lim_(x->+oo)3/(x+1)=0^+

The slant asymptote is y=2x-1

graph{(y-(2x^2+x+2)/(x+1))(y-2x+1)(y-50x-50)=0 [-25.64, 25.67, -12.83, 12.84]}