What is the minimum point of the parabola #y=2x^2-16x+5#?

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Answer 1 · 2016-08-19T14:46:45

The minimum is #y = -27#.

The minimum point will be the #y# coordinate of the vertex, or #q# in the form #y = a(x - p)^2 +q#.

Let's complete the square to transform into vertex form.

#y = 2(x^2 - 8x + n - n) + 5#

#n = (b/2)^2 = (-8/2)^2 = 16#

#y = 2(x^2 - 8x + 16 - 16) + 5#

#y = 2(x - 4)^2 - 16(2) + 5#

#y = 2(x - 4)^2 - 32 + 5#

#y = 2(x- 4)^2 - 27#

Hence, the vertex is at #(4, -27)#. So, the minimum is #y = -27#.

Hopefully this helps!