Question

How do you find the vertical, horizontal and slant asymptotes of: `g(x)= (x+2 )/( 2x^2)`?

Answer 1

The vertical asymptote is `x=0`
No slant asymptote
The horizontal asymptote is `y=0`

Explanation:

The domain of `g(x)` is `D_g(x)=RR-{0}`

As you cannot divide by `0`, `x!=0`

The vertical asymptote is `x=0`

The degree of the numerator is `<` than the degree of the numerator, there is no slant asymptote.

Now, we compute the limits of `g(x)` as `x->+-oo`

`lim_(x->-oo)g(x)=lim_(x->-oo)x/(2x^2)=lim_(x->-oo)1/(2x)=0^(-)`

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

The horizontal asymptote is `y=0`

graph{(x+2)/(2x^2) [-7.9, 7.9, -3.95, 3.95]}

Answer 2

vertical asymptote at x = 0
horizontal asymptote at y = 0

Explanation:

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to x and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : `2x^2=0rArrx^2=0rArrx=0`

`rArrx=0" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),g(x)toc" ( a constant)"`

divide terms on numerator/denominator by the highest power of x, that is `x^2`

`g(x)=(x/x^2+2/x^2)/((2x^2)/x^2)=(1/x+2/x^2)/2`

as `xto+-oo,g(x)to(0+0)/2`

`rArry=0" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x+2)/(2x^2) [-10, 10, -5, 5]}