SOLUTION: What is the minimum value of y = 2x^2-8x +11?
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Question 1058356
:
What is the minimum value of y = 2x^2-8x +11?
Found 2 solutions by
Alan3354, ewatrrr
:
Answer by
Alan3354(69443)
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What is the minimum value of y = 2x^2-8x +11?
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It's the vertex, on the Axis of Symmetry.
The AOS is x = -b/2a
x = 8/4 = 2
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y = 2*4 - 8*2 + 11
= 3
Answer by
ewatrrr(24785)
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y = 2x^2-8x +11
y = 2(x^2-4x) + 11
y = 2(x-2)^2 - 2*4 + 11
y = 2(x-2)^2 + 3
minimum value of y = 3