What are the discontinuities of the function f(x) = the quantity of x squared plus 6 x plus 9, all over 3 x plus 15.?

Factor: (x^2+6x+9)/(3x+15) = ((x+3)(x+3))/(3(x+5))

No factors cancel, so the discontinuity leads to a vertical asymptote. In this case, the vertical asymptote is x = -5 because this makes the denominator equal to zero. To find this value, set the denominator equal to 0 and solve for x

3(x+5) = 0
x+5 = 0
x = -5