titik potong sumbu x pada kurva f(x)=x2+ 10x +9 adalah.. tolong bantu ..

f(x) = 0
x^2 + 10x + 9 = 0
x^2 + x + 9x + 9 = 0
x(x+1) + 9(x+1) = 0
(x+1)(x+9) = 0
x+1 = 0 ==> x = -1
x+9 = 0==> x = -9

titik potong dgn sb X => (0,-1) dan (0,-9)