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## Homework Statement

For which values of b will the quadratic function ƒ(x) = x

^{2}-2bx+7 have a minimum value of 6?

## Homework Equations

y = ax

^{2}+bx+c

y = a(x-h)

^{2}+k

b(first one) = -2ah(second one)

c(first one) = ah

^{2}+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 = x
^{2}-2bx+7 - y = (x-h)
^{2}+6

^{2}-2hx+

__h__

^{2}+6∴ h

^{2}+6 = 7(c)

∴h = ±1

then because -2b = -2h,and h = ±1, b = ±1.

Is there a simpler way of doing this?

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