Find the Maximum/Minimum Value f(x)=7x^2-2x-8

$f (x) = 7 x^{2} - 2 x - 8$

Step 1

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 $x = - \frac{b}{2 a}$

Step 2

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

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

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

Remove parentheses.

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

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

Cancel the common factor of $- 2$ and $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{1}{7}$ .

Multiply $- 1$ by $- 1$ .

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

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

$x = \frac{1}{7}$

Step 3

Evaluate $f (\frac{1}{7})$ .

Replace the variable $x$ with $\frac{1}{7}$ in the expression.

$f (\frac{1}{7}) = 7 {(\frac{1}{7})}^{2} - 2 (\frac{1}{7}) - 8$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{7}$ .

$f (\frac{1}{7}) = 7 (\frac{1^{2}}{7^{2}}) - 2 (\frac{1}{7}) - 8$

One to any power is one.

$f (\frac{1}{7}) = 7 (\frac{1}{7^{2}}) - 2 (\frac{1}{7}) - 8$

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

$f (\frac{1}{7}) = 7 (\frac{1}{49}) - 2 (\frac{1}{7}) - 8$

Cancel the common factor of $7$ .

Factor $7$ out of $49$ .

$f (\frac{1}{7}) = 7 (\frac{1}{7 (7)}) - 2 (\frac{1}{7}) - 8$

Cancel the common factor.

$f (\frac{1}{7}) = 7 (\frac{1}{7 \cdot 7}) - 2 (\frac{1}{7}) - 8$

Rewrite the expression.

$f (\frac{1}{7}) = \frac{1}{7} - 2 (\frac{1}{7}) - 8$

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

$f (\frac{1}{7}) = \frac{1}{7} + \frac{- 2}{7} - 8$

Move the negative in front of the fraction.

$f (\frac{1}{7}) = \frac{1}{7} - \frac{2}{7} - 8$

Combine fractions.

Combine the numerators over the common denominator.

$f (\frac{1}{7}) = - 8 + \frac{1 - 2}{7}$

Simplify the expression.

Subtract $2$ from $1$ .

$f (\frac{1}{7}) = - 8 + \frac{- 1}{7}$

Move the negative in front of the fraction.

$f (\frac{1}{7}) = - 8 - \frac{1}{7}$

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$f (\frac{1}{7}) = - 8 \cdot \frac{7}{7} - \frac{1}{7}$

Combine $- 8$ and $\frac{7}{7}$ .

$f (\frac{1}{7}) = \frac{- 8 \cdot 7}{7} - \frac{1}{7}$

Combine the numerators over the common denominator.

$f (\frac{1}{7}) = \frac{- 8 \cdot 7 - 1}{7}$

Simplify the numerator.

Multiply $- 8$ by $7$ .

$f (\frac{1}{7}) = \frac{- 56 - 1}{7}$

Subtract $1$ from $- 56$ .

$f (\frac{1}{7}) = \frac{- 57}{7}$

Move the negative in front of the fraction.

$f (\frac{1}{7}) = - \frac{57}{7}$

The final answer is $- \frac{57}{7}$ .

$- \frac{57}{7}$

Step 4

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

$(\frac{1}{7}, - \frac{57}{7})$

Step 5

Find the Maximum/Minimum Value f(x)=7x^2-2x-8

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