Find the minimum value of $f (x) = x^{2} - 5 x + 6.$

\frac{1}{4}

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- \frac{3}{4}

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\frac{3}{4}

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- \frac{1}{4}

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Solution

The correct option is D $- \frac{1}{4}$
Minimum of $f (x) = - \frac{D}{4 a}$ at $x = - \frac{b}{2 a} = - (\frac{25 - 24}{4}) = \frac{- 1}{4}$ at $x = \frac{5}{2}$
Maximum of $f (x) \to \infty$
Hence range is $[- \frac{1}{4}, \infty) .$

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Find the minimum value of f(x)=x2−5x+6.

The correct option is D −14Minimum of f(x)=−D4a at x=−b2a=−(25−244)=−14 at x=52Maximum of f(x)→∞Hence range is [−14,∞).

Find the minimum value of $f (x) = x^{2} - 5 x + 6.$

The correct option is D $- \frac{1}{4}$
Minimum of $f (x) = - \frac{D}{4 a}$ at $x = - \frac{b}{2 a} = - (\frac{25 - 24}{4}) = \frac{- 1}{4}$ at $x = \frac{5}{2}$
Maximum of $f (x) \to \infty$
Hence range is $[- \frac{1}{4}, \infty) .$