Question 1000034
y=x^2 + bx + c has a = 1


"the vertex is given: (3,-4) "
(h,k) = (3,-4)
h = 3


h = -b/(2a)
3 = -b/(2*1) ... plug in h = 3 and a = 1
3 = -b/2
3*2 = -b
6 = -b
-b = 6
b = -6



y=x^2 + bx + c
y=x^2 - 6x + c ... plug in b = -6
-4=3^2 - 6(3) + c ... plug in (x,y) = (3,-4)
-4 = 9-18+c
-4 = -9+c
-4+9 = -9+c+9
5 = c
c = 5


So the equation is y = x^2 - 6x + 5