f(x)=x^2−12x+27
find Vertex (h, k)
x = h= -b/2a = -(-12)/2(1) = 6
k = f(h) = 36 - 12(6) + 27
k = -9
so vertex (6,-9)
y = a(x - h)^2 + k
so
y = (x - 6)^2 - 9
x intercept when y = 0
x^2−12x+27 = 0
(x - 9)(x - 3) = 0
x - 9 = 0; x = 9
x - 3 = 0; x = 3
so x-intercept (9,0); (3,0)
y intercept when x = 0 so y = 27 (0, 27)
OK now you can use these to graph the function
(6, -9) (9,0) and (3,0)
