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Precalculus Examples

f (x) = 3 x^{2} + 4

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

Step 1

The minimum of a quadratic function occurs at

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

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

a

$a$ is positive, the minimum value of the function is

f (- \frac{b}{2 a})

$f (- \frac{b}{2 a})$ .

f_{min}

$f_{min}$

x = a x^{2} + b x + c

$x = a x^{2} + b x + c$ occurs at

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

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

Step 2

Find the value of

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

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

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Step 2.1

Substitute in the values of

a

$a$ and

b

$b$ .

x = - \frac{0}{2 (3)}

$x = - \frac{0}{2 (3)}$

Step 2.2

Remove parentheses.

x = - \frac{0}{2 (3)}

$x = - \frac{0}{2 (3)}$

Step 2.3

Simplify

- \frac{0}{2 (3)}

$- \frac{0}{2 (3)}$ .

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Step 2.3.1

Cancel the common factor of

0

$0$ and

2

$2$ .

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Step 2.3.1.1

Factor

2

$2$ out of

0

$0$ .

x = - \frac{2 (0)}{2 (3)}

$x = - \frac{2 (0)}{2 (3)}$

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 3}

$x = - \frac{2 \cdot 0}{2 \cdot 3}$

Step 2.3.1.2.2

Rewrite the expression.

x = - \frac{0}{3}

$x = - \frac{0}{3}$

x = - \frac{0}{3}

$x = - \frac{0}{3}$

x = - \frac{0}{3}

$x = - \frac{0}{3}$

Step 2.3.2

Cancel the common factor of

0

$0$ and

3

$3$ .

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Step 2.3.2.1

Factor

3

$3$ out of

0

$0$ .

x = - \frac{3 (0)}{3}

$x = - \frac{3 (0)}{3}$

Step 2.3.2.2

Cancel the common factors.

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Step 2.3.2.2.1

Factor

3

$3$ out of

3

$3$ .

x = - \frac{3 \cdot 0}{3 \cdot 1}

$x = - \frac{3 \cdot 0}{3 \cdot 1}$

Step 2.3.2.2.2

Cancel the common factor.

x = - \frac{3 \cdot 0}{3 \cdot 1}

$x = - \frac{3 \cdot 0}{3 \cdot 1}$

Step 2.3.2.2.3

Rewrite the expression.

x = - \frac{0}{1}

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

Step 2.3.2.2.4

Divide

0

$0$ by

1

$1$ .

x = - 0

$x = - 0$

x = - 0

$x = - 0$

x = - 0

$x = - 0$

Step 2.3.3

Multiply

- 1

$- 1$ by

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

x = 0

$x = 0$

Step 3

Evaluate

f (0)

$f (0)$ .

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Step 3.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = 3 {(0)}^{2} + 4

$f (0) = 3 {(0)}^{2} + 4$

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

$0$ to any positive power yields

0

$0$ .

f (0) = 3 \cdot 0 + 4

$f (0) = 3 \cdot 0 + 4$

Step 3.2.1.2

Multiply

3

$3$ by

0

$0$ .

f (0) = 0 + 4

$f (0) = 0 + 4$

f (0) = 0 + 4

$f (0) = 0 + 4$

Step 3.2.2

Add

0

$0$ and

4

$4$ .

f (0) = 4

$f (0) = 4$

Step 3.2.3

The final answer is

4

$4$ .

4

$4$

4

$4$

4

$4$

Step 4

Use the

x

$x$ and

y

$y$ values to find where the minimum occurs.

(0, 4)

$(0, 4)$

Step 5

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

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