Find the Maximum/Minimum Value c(x)=x^2-20x+620

$c (x) = x^{2} - 20 x + 620$

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 $b$ .

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

Remove parentheses.

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

Simplify $- \frac{- 20}{2 (1)}$ .

Cancel the common factor of $- 20$ and $2$ .

Factor $2$ out of $- 20$ .

$x = - \frac{2 \cdot - 10}{2 \cdot 1}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 10}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 10}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{- 10}{1}$

Divide $- 10$ by $1$ .

$x = - - 10$

Multiply $- 1$ by $- 10$ .

$x = 10$

Step 3

Evaluate $f (10)$ .

Replace the variable $x$ with $10$ in the expression.

$c (10) = {(10)}^{2} - 20 \cdot 10 + 620$

Simplify the result.

Simplify each term.

Raise $10$ to the power of $2$ .

$c (10) = 100 - 20 \cdot 10 + 620$

Multiply $- 20$ by $10$ .

$c (10) = 100 - 200 + 620$

Simplify by adding and subtracting.

Subtract $200$ from $100$ .

$c (10) = - 100 + 620$

Add $- 100$ and $620$ .

$c (10) = 520$

The final answer is $520$ .

$520$

Step 4

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

$(10, 520)$

Step 5

Find the Maximum/Minimum Value c(x)=x^2-20x+620

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