Find the Maximum/Minimum Value f(x)=2x^2+4x+6

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

Step 1

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 2

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

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

$x = - \frac{4}{2 (2)}$

Remove parentheses.

$x = - \frac{4}{2 (2)}$

Simplify $- \frac{4}{2 (2)}$ .

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = 2 {(- 1)}^{2} + 4 (- 1) + 6$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = 2 \cdot 1 + 4 (- 1) + 6$

Multiply $2$ by $1$ .

$f (- 1) = 2 + 4 (- 1) + 6$

Multiply $4$ by $- 1$ .

$f (- 1) = 2 - 4 + 6$

Simplify by adding and subtracting.

Subtract $4$ from $2$ .

$f (- 1) = - 2 + 6$

Add $- 2$ and $6$ .

$f (- 1) = 4$

The final answer is $4$ .

$4$

Step 4

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

$(- 1, 4)$

Step 5

Find the Maximum/Minimum Value f(x)=2x^2+4x+6

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