Question

How do you identify all horizontal and slant asymptote for `f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2)`?

Answer 1

The vertical asymptote is `x=-2`
The slant asymptote is `y=2x-7`
There are no horizontal asymptote

Explanation:

The denominator can be factorised as `(x+2)(x+1)`
So the vertcal asymptote is `x=-2`
Let's do a long division to simplify `f(x)`
`color(white)(aaaa)` `2x^3-x^2-2x+1` `color(white)(aaaa)` `∣` `x^2+3x+2`
`color(white)(aaaa)` `2x^3+6x^2+4x` `color(white)(aaaaaaa)` `∣` `2x-7`
`color(white)(aaaaaa)` `0-7x^2-6x+1`
`color(white)(aaaaaaaa)` `-7x^2-21x-14`
`color(white)(aaaaaaaaa)` `-0+15x+15`

`(2x^3-x^2-2x+1)/(x^2+3x+2)=2x-7+(15cancel(x+1))/((x+2)cancel(x+1))`

`:. y=2x-7` is a slant asymptote
Also, `f(x)=((2x-1)(x+1)(x-1))/((x+2)(x+1))=((2x-1)(x-1))/(x+2)`
`lim_(x->+oo)f(x)=lim_(x->+oo)2x=+oo`
`lim_(x->-oo)f(x)=lim_(x->-oo)2x=-oo`

graph{(y-((2x^3-x^2-2x+1)/(x^2+3x+2)))(y-2x+7)=0 [-83.3, 83.4, -41.7, 41.74]}