Question

How do you find the asymptotes for `f(x) = x/((x+3)(x-4))`?

Answer 1

`"vertical asymptotes at "x=-3" and "x=4`
`"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 these values then they are vertical asymptotes.

`"solve "(x+3)(x-4)=0`

`rArrx=-3" and "x=4" 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)=(x/x^2)/(x^2/x^2-x/x^2-12/x^2)=(1/x)/(1-1/x-12/x^2)`

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

`y=0" is the asymptote"`
graph{x/((x+3)(x-4)) [-10, 10, -5, 5]}