Find the vertex of the parabola y=2x^2-16x+34

(4, 2)

Step-by-step explanation:

y = 2x^2 - 16x +34  can be simplified by factoring out 2 from all three terms on the right side:  y = 2(x^2 - 8x + 17).

We need to "complete the square" here.  Notice that the coefficient of x is -8; we take half of that (which is -4) and square this result (which yields 16).

Focusing on x^2 - 8x + 17, we add 16, and then subtract 16, obtaining:

                      x^2 - 8x + 16 - 16 + 17, or (x - 4)^2 + 1

Now return to the original function, y = 2(x^2 - 8x + 17).

Replace  x^2 - 8x + 17  with  [x - 4]^2 + 1:

We get     y = 2(x^2 - 8x + 17).   =    2(   [x - 4]^2 + 1  )

This has the form y = a(x - h)^2 + k, and by comparison we see that h = 4 and k = 2.  The vertex of this parabola is at  (4, 2).

hope this is what you were asking for!

step-by-step explanation:

something over 10 so ? /10

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