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

2 Answers
Jul 15, 2017

Answer:

#"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]}

Jul 15, 2017

Answer:

#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]}