Find the Maximum/Minimum Value f(x)=7x^2-7x

$f (x) = 7 x^{2} - 7 x$

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 $b$ .

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

Remove parentheses.

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

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

Cancel the common factor of $- 7$ and $7$ .

Factor $7$ out of $- 7$ .

$x = - \frac{7 \cdot - 1}{2 \cdot 7}$

Cancel the common factors.

Factor $7$ out of $2 \cdot 7$ .

$x = - \frac{7 \cdot - 1}{7 \cdot 2}$

Cancel the common factor.

$x = - \frac{7 \cdot - 1}{7 \cdot 2}$

Rewrite the expression.

$x = - \frac{- 1}{2}$

Move the negative in front of the fraction.

$x = - - \frac{1}{2}$

Multiply $- - \frac{1}{2}$ .

Multiply $- 1$ by $- 1$ .

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

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

$x = \frac{1}{2}$

Evaluate $f (\frac{1}{2})$ .

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

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

One to any power is one.

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

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

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

Combine $7$ and $\frac{1}{4}$ .

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

Combine $- 7$ and $\frac{1}{2}$ .

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

Move the negative in front of the fraction.

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

To write $- \frac{7}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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

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

Multiply $2$ by $2$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 7$ by $2$ .

$f (\frac{1}{2}) = \frac{7 - 14}{4}$

Subtract $14$ from $7$ .

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

Move the negative in front of the fraction.

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

The final answer is $- \frac{7}{4}$ .

$- \frac{7}{4}$

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

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

Find the Maximum/Minimum Value f(x)=7x^2-7x

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