Algebra Examples

$h (x) = - 2 x^{2} + 20 x + 48$

Step 1

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

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

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Step 2.1

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

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

Step 2.2

Remove parentheses.

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

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

Factor $2$ out of $20$ .

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

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

Cancel the common factor.

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

Step 2.3.1.2.3

Rewrite the expression.

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

Step 2.3.2

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

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Step 2.3.2.1

Factor $2$ out of $10$ .

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

Step 2.3.2.2

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$x = - (- 1 \cdot 5)$

Step 2.3.3

Multiply $- (- 1 \cdot 5)$ .

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Step 2.3.3.1

Multiply $- 1$ by $5$ .

$x = - - 5$

Step 2.3.3.2

Multiply $- 1$ by $- 5$ .

$x = 5$

Step 3

Evaluate $f (5)$ .

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Step 3.1

Replace the variable $x$ with $5$ in the expression.

$h (5) = - 2 {(5)}^{2} + 20 (5) + 48$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Raise $5$ to the power of $2$ .

$h (5) = - 2 \cdot 25 + 20 (5) + 48$

Step 3.2.1.2

Multiply $- 2$ by $25$ .

$h (5) = - 50 + 20 (5) + 48$

Step 3.2.1.3

Multiply $20$ by $5$ .

$h (5) = - 50 + 100 + 48$

Step 3.2.2

Simplify by adding numbers.

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Step 3.2.2.1

Add $- 50$ and $100$ .

$h (5) = 50 + 48$

Step 3.2.2.2

Add $50$ and $48$ .

$h (5) = 98$

Step 3.2.3

The final answer is $98$ .

$98$

Step 4

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

$(5, 98)$

Step 5