For what values of x, if any, does #f(x) = 1/((x+2)(x-1)) # have vertical asymptotes?
1 Answer
Feb 10, 2016
Explanation:
Vertical asymptotes will occur whenever the denominator of the function is equal to
So, the vertical asymptotes are at the values of
#(x+2)(x-1)=0#
Just like when solving a quadratic equation, we set both of these terms equal to
So, in setting both term equal to
#x+2=0" "=>" "color(red)(x=-2#
#x-1=0" "=>" "color(red)(x=1#
We can check a graph of the function:
graph{1/((x+2)(x-1)) [-5, 5, -5, 5]}
