Question

For what values of x, if any, does `f(x) = 1/((x^2-4)(x+3)(x-2))` have vertical asymptotes?

Answer 1

`"vertical asymptotes at "x=+-2" and "x=-3`

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

`"solve "(x^2-4)(x+3)(x-2)=0`

`rArr(x-2)^2(x+2)(x+3)=0`

`rArrx=+-2" and "x=-3" are the asymptotes"`
graph{1/((x^2-4)(x+3)(x-2)) [-10, 10, -5, 5]}

Answer 2

`x=-3,-2,2`

Explanation:

A vertical asymptote will occur when the denominator is zero.

We have:

`f(x) = 1/((x^2-4)(x+3)(x-2))`
`" " = 1/((x+2)(x-2)(x+3)(x-2))`
`" " = 1/((x+2)(x-2)^2(x+3))`

And so the denominator will be zero if:

`(x+2)(x-2)^2(x+3) = 0 => x=-3,-2,2`

Note that for `x=2` then as the factor in the denominator is `(x-2)^2` which is always positive then `f(x) rarr oo` irrespective of the sign of `x` as `x rarr 2`

We can confirm this graphically:

graph{1/((x^2-4)(x+3)(x-2)) [-6, 6, -5, 5]}