Algebra Examples

$- \frac{1}{3} x^{2} + \frac{4}{3} x - \frac{16}{3}$

Step 1

Simplify each term.

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Step 1.1

Combine $x^{2}$ and $\frac{1}{3}$ .

$f (x) = - \frac{x^{2}}{3} + \frac{4}{3} x - \frac{16}{3}$

Step 1.2

Combine $\frac{4}{3}$ and $x$ .

$f (x) = - \frac{x^{2}}{3} + \frac{4 x}{3} - \frac{16}{3}$

Step 2

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 3

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

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Step 3.1

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

$x = - \frac{1.\overline{3}}{2 (- 0.\overline{3})}$

Step 3.2

Remove parentheses.

$x = - \frac{1.\overline{3}}{2 (- 0.\overline{3})}$

Step 3.3

Simplify $- \frac{1.\overline{3}}{2 (- 0.\overline{3})}$ .

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Step 3.3.1

Multiply $2$ by $- 0.\overline{3}$ .

$x = - \frac{1.\overline{3}}{- 0.\overline{6}}$

Step 3.3.2

Divide $1.\overline{3}$ by $- 0.\overline{6}$ .

$x = - - 2$

Step 3.3.3

Multiply $- 1$ by $- 2$ .

$x = 2$

Step 4

Evaluate $f (2)$ .

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Step 4.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = - \frac{{(2)}^{2}}{3} + \frac{4 (2)}{3} - \frac{16}{3}$

Step 4.2

Simplify the result.

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Step 4.2.1

Combine the numerators over the common denominator.

$f (2) = \frac{- {(2)}^{2} + 4 (2) - 16}{3}$

Step 4.2.2

Simplify each term.

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Step 4.2.2.1

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

$f (2) = \frac{- 1 \cdot 4 + 4 (2) - 16}{3}$

Step 4.2.2.2

Multiply $- 1$ by $4$ .

$f (2) = \frac{- 4 + 4 (2) - 16}{3}$

Step 4.2.2.3

Multiply $4$ by $2$ .

$f (2) = \frac{- 4 + 8 - 16}{3}$

Step 4.2.3

Simplify the expression.

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Step 4.2.3.1

Add $- 4$ and $8$ .

$f (2) = \frac{4 - 16}{3}$

Step 4.2.3.2

Subtract $16$ from $4$ .

$f (2) = \frac{- 12}{3}$

Step 4.2.3.3

Divide $- 12$ by $3$ .

$f (2) = - 4$

Step 4.2.4

The final answer is $- 4$ .

$- 4$

Step 5

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

$(2, - 4)$

Step 6