For What Value Of B Does Y=x^2-2bx+9 Have A Minimum Value Of 8?

The extremum of the parabola is, where the 1st derivative is 0.

y = x^2 - 2bx + 9
y' = 2x - 2b

2x - 2b = 0
2x = 2b
x = b

1) This tells us that the parabola will have its extremum at x0 = b. (Btw, its a minimum, indeed, as y'' = 2 > 0.)

2) The minimum is supposed to be, where the value of this function is 8.

Let's represent these conditions in the equation of the parabola:

8 = x^2 - 2x^2 + 9 ;; -8 -x^2 +2x^2
x^2 = 1
x = 1

As x = b, b = 1.

And for b=1 your parabola has the equation:

y = x^2 - 2x + 9

That's all there is to it.
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Let's assume, the 2nd condition were:
2) The minimum is supposed to be, where the value of this function is 5.

5 = x^2 - 2x^2 + 9
x^2 = 4
x = 2

And, as before, as x = b, b = 2.
Then the parabola would have the equation:
y = x^2 - 4x + 9
with the minimum at x0 = 2.

Make a plot and verify.