Precalculus Examples

$g (x) = 2 x^{2} + 4 x + 3$

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

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Step 2.1

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

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

Step 2.2

Remove parentheses.

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

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

Factor $2$ out of $4$ .

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

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

Cancel the common factor.

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

Step 2.3.1.2.3

Rewrite the expression.

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

Step 2.3.2

Cancel the common factor of $2$ .

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Step 2.3.2.1

Cancel the common factor.

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

Step 2.3.2.2

Rewrite the expression.

$x = - 1 \cdot 1$

Step 2.3.3

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

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Step 3.1

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

$g (- 1) = 2 {(- 1)}^{2} + 4 (- 1) + 3$

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

$g (- 1) = 2 \cdot 1 + 4 (- 1) + 3$

Step 3.2.1.2

Multiply $2$ by $1$ .

$g (- 1) = 2 + 4 (- 1) + 3$

Step 3.2.1.3

Multiply $4$ by $- 1$ .

$g (- 1) = 2 - 4 + 3$

Step 3.2.2

Simplify by adding and subtracting.

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Step 3.2.2.1

Subtract $4$ from $2$ .

$g (- 1) = - 2 + 3$

Step 3.2.2.2

Add $- 2$ and $3$ .

$g (- 1) = 1$

Step 3.2.3

The final answer is $1$ .

$1$

Step 4

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

$(- 1, 1)$

Step 5