Question

How do you find the horizontal and vertical asymptote of the following: `f(x) = (2x-3)/(x^2+2)`?

Answer 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^2+2=0rArrx^2=-2`

`"this has no real solutions hence there are no vertical"`
`"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)/x^2-3/x^2)/(x^2/x^2+2/x^2)=(2/x-3/x^2)/(1+2/x^2)`

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

`rArry=0" is the asymptote"`
graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}