Find the Maximum/Minimum Value d(x)=1/2x^2-10x

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

Step 1

Combine $\frac{1}{2}$ and $x^{2}$ .

$d (x) = \frac{x^{2}}{2} - 10 x$

Step 2

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 3

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

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

$x = - \frac{- 10}{2 (0.5)}$

Remove parentheses.

$x = - \frac{- 10}{2 (0.5)}$

Simplify $- \frac{- 10}{2 (0.5)}$ .

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10$ .

$x = - \frac{2 \cdot - 5}{2 \cdot 0.5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 0.5$ .

$x = - \frac{2 \cdot - 5}{2 (0.5)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 5}{2 \cdot 0.5}$

Rewrite the expression.

$x = - \frac{- 5}{0.5}$

Simplify the expression.

Divide $- 5$ by $0.5$ .

$x = - - 10$

Multiply $- 1$ by $- 10$ .

$x = 10$

Step 4

Evaluate $f (10)$ .

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

$d (10) = \frac{{(10)}^{2}}{2} - 10 \cdot 10$

Simplify the result.

Simplify each term.

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

$d (10) = \frac{100}{2} - 10 \cdot 10$

Divide $100$ by $2$ .

$d (10) = 50 - 10 \cdot 10$

Multiply $- 10$ by $10$ .

$d (10) = 50 - 100$

Subtract $100$ from $50$ .

$d (10) = - 50$

The final answer is $- 50$ .

$- 50$

Step 5

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

$(10, - 50)$

Step 6

Find the Maximum/Minimum Value d(x)=1/2x^2-10x

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