Find the Maximum/Minimum Value c(x)=x^2-40x+680

$c (x) = x^{2} - 40 x + 680$

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 $b$ .

$x = - \frac{- 40}{2 (1)}$

Remove parentheses.

$x = - \frac{- 40}{2 (1)}$

Simplify $- \frac{- 40}{2 (1)}$ .

Cancel the common factor of $- 40$ and $2$ .

Factor $2$ out of $- 40$ .

$x = - \frac{2 \cdot - 20}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$x = - \frac{2 \cdot - 20}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 20}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{- 20}{1}$

Divide $- 20$ by $1$ .

$x = - - 20$

Multiply $- 1$ by $- 20$ .

$x = 20$

Step 3

Evaluate $f (20)$ .

Replace the variable $x$ with $20$ in the expression.

$c (20) = {(20)}^{2} - 40 \cdot 20 + 680$

Simplify the result.

Simplify each term.

Raise $20$ to the power of $2$ .

$c (20) = 400 - 40 \cdot 20 + 680$

Multiply $- 40$ by $20$ .

$c (20) = 400 - 800 + 680$

Simplify by adding and subtracting.

Subtract $800$ from $400$ .

$c (20) = - 400 + 680$

Add $- 400$ and $680$ .

$c (20) = 280$

The final answer is $280$ .

$280$

Step 4

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

$(20, 280)$

Step 5

Find the Maximum/Minimum Value c(x)=x^2-40x+680

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