Question

Question #16c31

Answer 1

The vertical asymptotes are `x=0` and `x=-2`
The horizontal asymptote is `y=0`
No oblique asymptote

Explanation:

We factorise the denominator

`2x^3+4x^2=2x^2(x+2)`

As you cannot divide by `0`, `x!=0` and `x!=-2`

The vertical asymptotes are `x=0` and `x=-2`

The degree of the numerator is `<` than the degree of the denominator, there is no oblique asymptote,

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

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

The horizontal asymptote is `y=0`

graph{(y-(1-2x^2)/(2x^3+4x^2))(y)(y-100x)(y-100x-200)=0 [-4.935, 4.93, -2.46, 2.474]}