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

1 Answer
Apr 24, 2016

Answer:

vertical asymptotes are x = -3 and x = 2

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : (x+3)(x-2) = 0 → x = -3 , x = 2

#rArr x = -3" and " x = 2" are the asymptotes " #
graph{1/((x+3)(x-2)) [-10, 10, -5, 5]}