Question

Question #3a7aa

Answer 1

`"vertical asymptotes at "x=+-2`
`"horizontal asymptote at "y=-1`

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 values x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

`"solve "4-x^2=0rArr(2-x)(2+x)=0`

`rArrx=-2" and "x=2" are the asymptotes"`

`"horizontal asymptotes occur as"`

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

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

`f(x)=(x^2/x^2)/(4/x^2-x^2/x^2)=1/(4/x^2-1)`

`"as "xto+-oo,f(x)to1/(0-1)`

`rArry=-1" is the asymptote"`
graph{x^2/(4-x^2) [-10, 10, -5, 5]}