What value of x will give the minimum value of $x^{2}$ - 4x + 7?

\geq

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x = 2

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x < 2

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x = 0

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Solution

The correct option is B x = 2
Suppose y = $x^{2}$ - 4x + 7
For minimum value of y, $\frac{d y}{d x}$ must be 0.
$= \frac{d}{d x} (x^{2}) - 4 \frac{d (x)}{d x} + \frac{d (7)}{d x} = 0$
2x - 4 + 0 = 0
2x = 4
x = 2
Alternate Method:
For minimum, value of x $= \frac{- b}{2 a}$
$∴ x = \frac{4}{2 \times 1} = 2$

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