Question

How do you find the asymptotes for `f(x)=(2x^3+3x^2+x)/(6x^2+x-1)`?

Answer 1

The vertical asymptote is `x=1/3`. No horizontal asymptote. The slant asymptote is `y=1/3x+4/9`

Explanation:

The numerator is

`2x^3+3x^2+x=x(2x^2+3x+1)`

`=x(2x+1)(x+1)`

The denominator is

`6x^2+x-1=(3x-1)(2x+1)`

The function is

`f(x)=(2x^3+3x^2+x)/(6x^2+x-1)`

`=(xcancel(2x+1)(x+1))/((3x-1)cancel(2x+1))`

`=(x(x+1))/(3x-1)`

As the denominator must de `!=0`, therefore

`3x-1!=0`, `=>`, `x!=1/3`

The vertical asymptote is `x=1/3`

There is no horizontal asymptote, as

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

`lim_(x->-oo)f(x)=lim_(x->-oo)(x^2)/(3x)=lim_(x->-oo)x/3=-oo`

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

Perform a long division

`color(white)(aaaa)` `x^2+x` `color(white)(aaaa)` `|` `3x-1`

`color(white)(aaaa)` `x^2-1/3x` `color(white)(aaaa)` `|` `1/3x+4/9`

`color(white)(aaaaa)` `0+4/3x`

`color(white)(aaaaaaa)` `+4/3x+0`

`color(white)(aaaaaaa)` `+4/3x-4/9`

`color(white)(aaaaaaaaa)` `+0+4/9`

Therefore,

`f(x)=(x^2+x)/(3x-1)=(1/3x+4/9)+(4/9)/(3x-1)`

`lim_(x->+oo)(f(x)-(1/3x+4/9))=lim_(x->+oo)(4/9)/(3x-1)=0^+`

`lim_(x->-oo)(f(x)-(1/3x+4/9))=lim_(x->-oo)(4/9)/(3x-1)=0^-`

The slant asymptote is

`y=1/3x+4/9`

graph{(y-(x(x+1))/(3x-1))(y-1/3x-4/9)=0 [-4.382, 4.386, -2.19, 2.193]}