Precalculus Examples

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

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

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Step 2.1

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

$x = - \frac{0}{2 (5)}$

Step 2.2

Remove parentheses.

$x = - \frac{0}{2 (5)}$

Step 2.3

Simplify $- \frac{0}{2 (5)}$ .

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Step 2.3.1

Cancel the common factor of $0$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $0$ .

$x = - \frac{2 (0)}{2 (5)}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Cancel the common factor.

$x = - \frac{2 \cdot 0}{2 \cdot 5}$

Step 2.3.1.2.2

Rewrite the expression.

$x = - \frac{0}{5}$

Step 2.3.2

Cancel the common factor of $0$ and $5$ .

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Step 2.3.2.1

Factor $5$ out of $0$ .

$x = - \frac{5 (0)}{5}$

Step 2.3.2.2

Cancel the common factors.

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Step 2.3.2.2.1

Factor $5$ out of $5$ .

$x = - \frac{5 \cdot 0}{5 \cdot 1}$

Step 2.3.2.2.2

Cancel the common factor.

$x = - \frac{5 \cdot 0}{5 \cdot 1}$

Step 2.3.2.2.3

Rewrite the expression.

$x = - \frac{0}{1}$

Step 2.3.2.2.4

Divide $0$ by $1$ .

$x = - 0$

Step 2.3.3

Multiply $- 1$ by $0$ .

$x = 0$

Step 3

Evaluate $f (0)$ .

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Step 3.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 5 {(0)}^{2} + 6$

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

Raising $0$ to any positive power yields $0$ .

$f (0) = 5 \cdot 0 + 6$

Step 3.2.1.2

Multiply $5$ by $0$ .

$f (0) = 0 + 6$

Step 3.2.2

Add $0$ and $6$ .

$f (0) = 6$

Step 3.2.3

The final answer is $6$ .

$6$

Step 4

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

$(0, 6)$

Step 5

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