SOLUTION: Express x^2-5x+6 in the form of (x-a)^2-b. Hence state the coordinates of the turning point of the curve y=x^2-5x+6. I really don't understand how to work out the turning poin

Algebra ->  Graphs -> SOLUTION: Express x^2-5x+6 in the form of (x-a)^2-b. Hence state the coordinates of the turning point of the curve y=x^2-5x+6. I really don't understand how to work out the turning poin      Log On


   

Question 919341: Express x^2-5x+6 in the form of (x-a)^2-b. Hence state the coordinates of the turning point of the curve y=x^2-5x+6.

I really don't understand how to work out the turning point so guidance would be very helpful. Thanks.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
the vertex form of a Parabola opening up(a>0) or down(a<0),
y=a%28x-h%29%5E2+%2Bk. V(h, k)
Completing the Square to Obtain the Vertex Form:
y = ax^2 + bx + c = 0
y = x^2-5x+6 = 0 b%2F%28-2a%29+=+%28-5%29%2F%28-2%29+=+5%2F2 5/2 the x-value for the Vertex
y = (x - 5/2)^2 - (5/2)^2 + 6 = 0
y = (x - 5/2)^2 - 25/4 + 24/4= 0 -a%28b%2F%28-2a%29%29%5E2+= -25/4
y = (x - 5/2)^2 - 1/4 = 0
V(5/2, -1/4)