Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)=(-2r^2+5) /( r^2 - 4)`?

Answer 1

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

Explanation:

`f(x)=(-2x^2+5)/(x^2-4)`

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

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

`rArrx=+-2" are the asymptotes"`

`"horizontal asymptotes occur as"`

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

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

`f(x)=((-2x^2)/x^2+5/x^2)/(x^2/x^2-4/x^2)=(-2+5/x^2)/(1-4/x^2)`

as `xto+-oo,f(x)to(-2+0)/(1-0)`

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