Find the Maximum/Minimum Value (x+3)^2-5

${(x + 3)}^{2} - 5$

Simplify each term.

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

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

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$f (x) = x \cdot x + x \cdot 3 + 3 (x + 3) - 5$

Apply the distributive property.

$f (x) = x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3 - 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $3$ to the left of $x$ .

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

Multiply $3$ by $3$ .

$f (x) = x^{2} + 3 x + 3 x + 9 - 5$

Add $3 x$ and $3 x$ .

$f (x) = x^{2} + 6 x + 9 - 5$

Subtract $5$ from $9$ .

$f (x) = x^{2} + 6 x + 4$

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 $x = - \frac{b}{2 a}$

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

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

$x = - \frac{6}{2 (1)}$

Remove parentheses.

$x = - \frac{6}{2 (1)}$

Simplify $- \frac{6}{2 (1)}$ .

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot 1}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 3}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot 3}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{3}{1}$

Divide $3$ by $1$ .

$x = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x = - 3$

Evaluate $f (- 3)$ .

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

$f (- 3) = {(- 3)}^{2} + 6 (- 3) + 4$

Simplify the result.

Simplify each term.

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

$f (- 3) = 9 + 6 (- 3) + 4$

Multiply $6$ by $- 3$ .

$f (- 3) = 9 - 18 + 4$

Simplify by adding and subtracting.

Subtract $18$ from $9$ .

$f (- 3) = - 9 + 4$

Add $- 9$ and $4$ .

$f (- 3) = - 5$

The final answer is $- 5$ .

$- 5$

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

$(- 3, - 5)$

Find the Maximum/Minimum Value (x+3)^2-5

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