Question

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

Answer 1

vertical asymptote at `x=-1/3`
horizontal asymptote at `y=2/9`

Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero 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: `(-3x-1)^2=0rArr-3x-1=0rArrx=-1/3`

`rArrx=-1/3" is the asymptote"`

Horizontal asymptotes occur as

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

`f(x)=(2x^2)/((-3x-1)^2)=(2x^2)/(9x^2+6x+1)`

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

`f(x)=((2x^2)/(x^2))/((9x^2)/x^2+(6x)/x^2+1/x^2)=2/(9+6/x+1/x^2)`

as `xto+-oo,f(x)to2/(9+0+0)`

`rArry=2/9" is the asymptote"`
graph{(2x^2)/(9x^2+6x+1) [-20, 20, -10, 10]}