Precalculus Examples

$3 x^{2} - 6 x + 13$

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 $b$ .

$x = - \frac{- 6}{2 (3)}$

Step 2.2

Remove parentheses.

$x = - \frac{- 6}{2 (3)}$

Step 2.3

Simplify $- \frac{- 6}{2 (3)}$ .

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Step 2.3.1

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

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Step 2.3.1.1

Factor $2$ out of $- 6$ .

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

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor $2$ out of $2 \cdot 3$ .

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

Step 2.3.1.2.2

Cancel the common factor.

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

Step 2.3.1.2.3

Rewrite the expression.

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

Step 2.3.2

Cancel the common factor of $- 3$ and $3$ .

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Step 2.3.2.1

Factor $3$ out of $- 3$ .

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

Step 2.3.2.2

Cancel the common factors.

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Step 2.3.2.2.1

Factor $3$ out of $3$ .

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

Step 2.3.2.2.2

Cancel the common factor.

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

Step 2.3.2.2.3

Rewrite the expression.

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

Step 2.3.2.2.4

Divide $- 1$ by $1$ .

$x = - - 1$

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.

$f (1) = 3 {(1)}^{2} - 6 \cdot 1 + 13$

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

One to any power is one.

$f (1) = 3 \cdot 1 - 6 \cdot 1 + 13$

Step 3.2.1.2

Multiply $3$ by $1$ .

$f (1) = 3 - 6 \cdot 1 + 13$

Step 3.2.1.3

Multiply $- 6$ by $1$ .

$f (1) = 3 - 6 + 13$

Step 3.2.2

Simplify by adding and subtracting.

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Step 3.2.2.1

Subtract $6$ from $3$ .

$f (1) = - 3 + 13$

Step 3.2.2.2

Add $- 3$ and $13$ .

$f (1) = 10$

Step 3.2.3

The final answer is $10$ .

$10$

Step 4

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

$(1, 10)$

Step 5