Precalculus Examples

$f (x) = x^{2} + 2 x - 15$

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 $b$ .

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

Step 2.2

Remove parentheses.

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

Step 2.3

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

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Step 2.3.1

Cancel the common factor of $2$ .

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Step 2.3.1.1

Cancel the common factor.

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

Step 2.3.1.2

Rewrite the expression.

$x = - 1 \cdot 1$

Step 2.3.2

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.

$f (- 1) = {(- 1)}^{2} + 2 (- 1) - 15$

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 $2$ .

$f (- 1) = 1 + 2 (- 1) - 15$

Step 3.2.1.2

Multiply $2$ by $- 1$ .

$f (- 1) = 1 - 2 - 15$

Step 3.2.2

Simplify by subtracting numbers.

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Step 3.2.2.1

Subtract $2$ from $1$ .

$f (- 1) = - 1 - 15$

Step 3.2.2.2

Subtract $15$ from $- 1$ .

$f (- 1) = - 16$

Step 3.2.3

The final answer is $- 16$ .

$- 16$

Step 4

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

$(- 1, - 16)$

Step 5