Find the Maximum/Minimum Value y=1/3x^2+2x+5

$y = \frac{1}{3} x^{2} + 2 x + 5$

Step 1

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

$f (x) = \frac{x^{2}}{3} + 2 x + 5$

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 $b$ .

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

Remove parentheses.

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

$x = - \frac{2}{2 \cdot 0.\overline{3}}$

Rewrite the expression.

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

Simplify the expression.

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

$x = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x = - 3$

Step 4

Evaluate $f (- 3)$ .

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = \frac{{(- 3)}^{2}}{3} + 2 (- 3) + 5$

Simplify the result.

Simplify each term.

Raise $- 3$ to the power of $2$ .

$f (- 3) = \frac{9}{3} + 2 (- 3) + 5$

Divide $9$ by $3$ .

$f (- 3) = 3 + 2 (- 3) + 5$

Multiply $2$ by $- 3$ .

$f (- 3) = 3 - 6 + 5$

Simplify by adding and subtracting.

Subtract $6$ from $3$ .

$f (- 3) = - 3 + 5$

Add $- 3$ and $5$ .

$f (- 3) = 2$

The final answer is $2$ .

$2$

Step 5

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

$(- 3, 2)$

Step 6

Find the Maximum/Minimum Value y=1/3x^2+2x+5

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