Question

How do you identify all asymptotes or holes for `f(x)=(x^3+3x^2+2x)/(3x^2+15x+12)`?

Answer 1

There is a hole at `x=-1`
A vertical asymptote is `(x=-4)`

The slant asymptote is `y=x/3-2/3`
No horizontal asymptote

Explanation:

Let's simplify `f(x)`

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

The denominator `=3x^2+15x+12=3(x^2+5x+4)`
`=3(x+1)(x+4)`

Therefore, `f(x)=(xcancel(x+1)(x+2))/(3cancel(x+1)(x+4))`

There is a hole when `x=-1`

`:. f(x)=(x(x+2))/(3(x+4))`

As we cannot divide by `0`, `x!=-4`

So, a vertical asymptote is `(x=-4)`

As the degree of the numerator `>` degree of the denominator, we expect a slant asymptote.

Let `y=ax+b` be the slant asymptote

`f(x)=(x^2+2x)/(3x+12)=(ax+b)+c/(3x+12)`

`=((ax+b)(3x+12)+c)/(3x+12)`

`:.x^2+2x=(ax+b)(3x+12)+c`

Comparing the coefficients,

of `x^2` `=>`, `3a=1`, `a=1/3`

of `x,` `=>`, `2=3b+12a`, `=>` `b=-2/3`

and `12b+c=0` `=>`, `c=8`

The slant asymptote is `y=x/3-2/3`

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

Therefore, no horizontal asymtote

graph{(y-(x^2+2x)/(3x+12))(y-x/3+2/3)=0 [-10, 10, -5, 5]}