Question

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

Answer 1

`"vertical asymptote at "x=-1`
`"horizontal asymptote at "y=0`

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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 "(x+1)^2=0rArrx=-1" is the asymptote"`

`"horizontal asymptotes occur as"`

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

`"divide terms on numerator/denominator by "x^2`

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

as `xto+-oo,f(x)to0/(1+0+0)`

`rArry=0" is the asymptote"`
graph{-1/(x+1)^2 [-10, 10, -5, 5]}