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What is the minimum value for the function y=2x^2-32x+256

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The answer would be: 128

You can find minimum value by looking at the vertex. First, differentiate the equation and find the value of x.

y=2x^2-32x+256
4x - 32=0
4x=32
x=8

If you put x=8, the value would be:
y=2x^2-32x+256
y=2(8)^2-32(8)+256
y= 128- 256 + 256= 128
User Gayan Jayasingha
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