How do you identify all asymptotes or holes for $f(x)=(x^2+11x+18)/(2x+1)$ ?

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Answer 1 · 2016-12-16T06:31:04

There is no hole. We have vertical asymptote at $x=-1/2$ and oblique asymptote at $2x-4y+21=0$

Let us examine $f(x)=(x^2+11x+18)/(2x+1)$ .

As $x^2+11x+18=x^2+9x+2x+18$

= $x(x+9)+2(x+9)=(x+2)(x+9)$

As a common factor cannot be crossed out from both the numerator and the denominators, we do not have a hole in $f(x)$ .

It is also apparent that as $x->-1/2$ , $(2x+1)->0$

and as we have $f(x)=(x^2+11x+18)/(2x+1)$ , $f(x)->oo$

Hence, we have a vertical asymptote at $x=-1/2$ .

Further, $f(x)=(x^2+11x+18)/(2x+1)$

= $(x/2(2x+1)+21/4(2x+1)+51/4)/(2x+1)$

= $x/2+21/4+51/(4(2x+1)$

Hence, as $x->oo$ . $f(x)->x/2+21/4$

and we have an oblique asymptote at $y=x/2+21/4$ or $2x-4y+21=0$
graph{(y-(x^2+11x+18)/(2x+1))(x+1/2)(2x-4y+21)=0 [-21.08, 18.92, -5.32, 14.68]}

How do you identify all asymptotes or holes for f(x)=(x^2+11x+18)/(2x+1)f(x)=(x^2+11x+18)/(2x+1)?