Find b and c so that y=18x^2+bx+c has vertex (-3,10)

Answer 1

y=a(x-h)^2 + k where (h,k) = vertex = (-3,10)

y=a(x+3)^2 +10 = a(x^2 +6x +9) +10 = ax^2 +6ax + 9a+10

a=18,

b =6a =6(18) = 108

c =9a+10 = 9(18)+10 = 162+10 = 172

b= 108, c=172