For what values of x, if any, does #f(x) = 1/((x-4)(x+8)) # have vertical asymptotes?

1 Answer
Jan 2, 2016

Answer:

#x=-8,4#

Explanation:

Vertical asymptotes occur at breaks in the domain of a function. This happens when plugging in a value for #x# would cause an issue.

The only such issue that could occur in this function is dividing a number by #0#.

Here, if the denominator equals #0#, the graph will have a vertical asymptote.

Thus, to find the spots where the graph has vertical asymptotes, set the denominator of the function equal to #0#.

#(x-4)(x+8)=0#

Just like in working with quadratics, recall that either one of the two parts being multiplied can be equal to #0#.

#{(x-4=0),(x+8=0):}#

Solve for both.

#{(x=4),(x=-8):}#

Vertical asymptotes occur at #x=-8,4#.

graph{1/((x-4)(x+8)) [-14.51, 10.81, -5.22, 7.44]}