For which values of b will the quadratic function ƒ(x) = x2-2bx+7 have a minimum value of 6?
y = ax2+bx+c
y = a(x-h)2+k
b(first one) = -2ah(second one)
c(first one) = ah2+k(second one)
The Attempt at a Solution
Well, I actually already have the answer to the question: b = ±1, but I believe there must be a better method to this.
My poor method goes like this:
Since the parabola has a "minimum", it must open upwards,with a vertex's y position as 6, which makes the function I am dealing with looks like these two forms:
- y = x2-2bx+7
- y = (x-h)2+6
∴ h2+6 = 7(c)
∴h = ±1
then because -2b = -2h,and h = ±1, b = ±1.
Is there a simpler way of doing this?