Find minimum value of y in $y = 4 x^{2} - 4 x + 7$

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Solution

$y = 4 x^{2} - 4 x + 7 y^{'} = 8 x - 4 = 0 x = \frac{1}{2} y^{''} = 8 > 0 S o y_{m i n} = 4. \frac{1}{4} - 4 \frac{1}{2} + 7 y_{m i n} = 6$

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