Question

find the vertex of the parabola: f(x)= x^2-x+1

Answer 1

a=1
b=-1
c=1
x = -b/2a
x= -(-1)/2(1) = 1/2
y = (1/2)^2 - 1/2 +1
y = 1/4 - 1/2 +1
y = 3/4
vertex = (1/2, 3/4)

Also you could use completing the square to put it in vertex form
y = a(x-h)^2 +k where vertex = (h,k)
take half of b
b/2 = -1/2
this will be the "-h" part
now square it and subtract from c
(-1/2)^2 = 1/4
1 - 1/4 = 3/4
y = (x- 1/2)^2 + 3/4
vertex = (1/2, 3/4)