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Finite Math Examples

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

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

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.

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

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

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

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

$1$ .

$x = - 0$

$x = - 0$

$x = - 0$

Step 2.3.3

Multiply

$- 1$ by

$0$ .

$x = 0$

$x = 0$

$x = 0$

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

$0$ .

$f (0) = 0 + 6$

$f (0) = 0 + 6$

Step 3.2.2

Add

$0$ and

$6$ .

$f (0) = 6$

Step 3.2.3

The final answer is

$6$ .

$6$

$6$

$6$

Step 4

Use the

$x$ and

$y$ values to find where the minimum occurs.

$(0, 6)$

Step 5

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$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$