How do you find the range of $y=4x²-5x+2$ ?

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Answer 1 · 2018-06-10T09:30:28

$y>=7/16$

We get

$y'(x)=8x-5$
$y''(x)=8$
So we get

$y(5/8)=4*25/64-25/8+2=7/16$

Since we have for $x=5/8$ a minimum, so we get

$y>=7/16$

How do you find the range of y=4x²-5x+2y=4x²-5x+2?