Find the Maximum/Minimum Value x^2-7x+12

$x^{2} - 7 x + 12$

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{- 7}{2 (1)}$

Remove parentheses.

$x = - \frac{- 7}{2 (1)}$

Simplify $- \frac{- 7}{2 (1)}$ .

Multiply $2$ by $1$ .

$x = - \frac{- 7}{2}$

Move the negative in front of the fraction.

$x = - - \frac{7}{2}$

Multiply $- - \frac{7}{2}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{7}{2})$

Multiply $\frac{7}{2}$ by $1$ .

$x = \frac{7}{2}$

Evaluate $f (\frac{7}{2})$ .

Replace the variable $x$ with $\frac{7}{2}$ in the expression.

$f (\frac{7}{2}) = {(\frac{7}{2})}^{2} - 7 (\frac{7}{2}) + 12$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{7}{2}$ .

$f (\frac{7}{2}) = \frac{7^{2}}{2^{2}} - 7 (\frac{7}{2}) + 12$

Raise $7$ to the power of $2$ .

$f (\frac{7}{2}) = \frac{49}{2^{2}} - 7 (\frac{7}{2}) + 12$

Raise $2$ to the power of $2$ .

$f (\frac{7}{2}) = \frac{49}{4} - 7 (\frac{7}{2}) + 12$

Multiply $- 7 (\frac{7}{2})$ .

Combine $- 7$ and $\frac{7}{2}$ .

$f (\frac{7}{2}) = \frac{49}{4} + \frac{- 7 \cdot 7}{2} + 12$

Multiply $- 7$ by $7$ .

$f (\frac{7}{2}) = \frac{49}{4} + \frac{- 49}{2} + 12$

Move the negative in front of the fraction.

$f (\frac{7}{2}) = \frac{49}{4} - \frac{49}{2} + 12$

Find the common denominator.

Multiply $\frac{49}{2}$ by $\frac{2}{2}$ .

$f (\frac{7}{2}) = \frac{49}{4} - (\frac{49}{2} \cdot \frac{2}{2}) + 12$

Multiply $\frac{49}{2}$ by $\frac{2}{2}$ .

$f (\frac{7}{2}) = \frac{49}{4} - \frac{49 \cdot 2}{2 \cdot 2} + 12$

Write $12$ as a fraction with denominator $1$ .

$f (\frac{7}{2}) = \frac{49}{4} - \frac{49 \cdot 2}{2 \cdot 2} + \frac{12}{1}$

Multiply $\frac{12}{1}$ by $\frac{4}{4}$ .

$f (\frac{7}{2}) = \frac{49}{4} - \frac{49 \cdot 2}{2 \cdot 2} + \frac{12}{1} \cdot \frac{4}{4}$

Multiply $\frac{12}{1}$ by $\frac{4}{4}$ .

$f (\frac{7}{2}) = \frac{49}{4} - \frac{49 \cdot 2}{2 \cdot 2} + \frac{12 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (\frac{7}{2}) = \frac{49}{4} - \frac{49 \cdot 2}{4} + \frac{12 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (\frac{7}{2}) = \frac{49 - 49 \cdot 2 + 12 \cdot 4}{4}$

Simplify each term.

Multiply $- 49$ by $2$ .

$f (\frac{7}{2}) = \frac{49 - 98 + 12 \cdot 4}{4}$

Multiply $12$ by $4$ .

$f (\frac{7}{2}) = \frac{49 - 98 + 48}{4}$

Simplify the expression.

Subtract $98$ from $49$ .

$f (\frac{7}{2}) = \frac{- 49 + 48}{4}$

Add $- 49$ and $48$ .

$f (\frac{7}{2}) = \frac{- 1}{4}$

Move the negative in front of the fraction.

$f (\frac{7}{2}) = - \frac{1}{4}$

The final answer is $- \frac{1}{4}$ .

$- \frac{1}{4}$

Use the $x$ and $y$ values to find where the minimum occurs.

$(\frac{7}{2}, - \frac{1}{4})$

Find the Maximum/Minimum Value x^2-7x+12

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