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Precalculus Examples

f (x) = 5 x^{2} - 2

$f (x) = 5 x^{2} - 2$

Step 1

Rewrite the equation in vertex form.

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Step 1.1

Complete the square for

5 x^{2} - 2

$5 x^{2} - 2$ .

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Step 1.1.1

Use the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , to find the values of

a

$a$ ,

b

$b$ , and

c

$c$ .

a = 5

$a = 5$

b = 0

$b = 0$

c = - 2

$c = - 2$

Step 1.1.2

Consider the vertex form of a parabola.

a {(x + d)}^{2} + e

$a {(x + d)}^{2} + e$

Step 1.1.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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Step 1.1.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

d = \frac{0}{2 \cdot 5}

$d = \frac{0}{2 \cdot 5}$

Step 1.1.3.2

Simplify the right side.

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Step 1.1.3.2.1

Cancel the common factor of

0

$0$ and

2

$2$ .

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Step 1.1.3.2.1.1

Factor

2

$2$ out of

0

$0$ .

d = \frac{2 (0)}{2 \cdot 5}

$d = \frac{2 (0)}{2 \cdot 5}$

Step 1.1.3.2.1.2

Cancel the common factors.

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Step 1.1.3.2.1.2.1

Factor

2

$2$ out of

2 \cdot 5

$2 \cdot 5$ .

d = \frac{2 (0)}{2 (5)}

$d = \frac{2 (0)}{2 (5)}$

Step 1.1.3.2.1.2.2

Cancel the common factor.

d = \frac{2 \cdot 0}{2 \cdot 5}

$d = \frac{2 \cdot 0}{2 \cdot 5}$

Step 1.1.3.2.1.2.3

Rewrite the expression.

d = \frac{0}{5}

$d = \frac{0}{5}$

d = \frac{0}{5}

$d = \frac{0}{5}$

d = \frac{0}{5}

$d = \frac{0}{5}$

Step 1.1.3.2.2

Cancel the common factor of

0

$0$ and

5

$5$ .

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Step 1.1.3.2.2.1

Factor

5

$5$ out of

0

$0$ .

d = \frac{5 (0)}{5}

$d = \frac{5 (0)}{5}$

Step 1.1.3.2.2.2

Cancel the common factors.

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Step 1.1.3.2.2.2.1

Factor

5

$5$ out of

5

$5$ .

d = \frac{5 \cdot 0}{5 \cdot 1}

$d = \frac{5 \cdot 0}{5 \cdot 1}$

Step 1.1.3.2.2.2.2

Cancel the common factor.

d = \frac{5 \cdot 0}{5 \cdot 1}

$d = \frac{5 \cdot 0}{5 \cdot 1}$

Step 1.1.3.2.2.2.3

Rewrite the expression.

d = \frac{0}{1}

$d = \frac{0}{1}$

Step 1.1.3.2.2.2.4

Divide

0

$0$ by

1

$1$ .

d = 0

$d = 0$

d = 0

$d = 0$

d = 0

$d = 0$

d = 0

$d = 0$

d = 0

$d = 0$

Step 1.1.4

Find the value of

e

$e$ using the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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Step 1.1.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

e = - 2 - \frac{0^{2}}{4 \cdot 5}

$e = - 2 - \frac{0^{2}}{4 \cdot 5}$

Step 1.1.4.2

Simplify the right side.

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Step 1.1.4.2.1

Simplify each term.

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Step 1.1.4.2.1.1

Raising

0

$0$ to any positive power yields

0

$0$ .

e = - 2 - \frac{0}{4 \cdot 5}

$e = - 2 - \frac{0}{4 \cdot 5}$

Step 1.1.4.2.1.2

Multiply

4

$4$ by

5

$5$ .

e = - 2 - \frac{0}{20}

$e = - 2 - \frac{0}{20}$

Step 1.1.4.2.1.3

Divide

0

$0$ by

20

$20$ .

e = - 2 - 0

$e = - 2 - 0$

Step 1.1.4.2.1.4

Multiply

- 1

$- 1$ by

0

$0$ .

e = - 2 + 0

$e = - 2 + 0$

e = - 2 + 0

$e = - 2 + 0$

Step 1.1.4.2.2

Add

- 2

$- 2$ and

0

$0$ .

e = - 2

$e = - 2$

e = - 2

$e = - 2$

e = - 2

$e = - 2$

Step 1.1.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

5 {(x + 0)}^{2} - 2

$5 {(x + 0)}^{2} - 2$ .

5 {(x + 0)}^{2} - 2

$5 {(x + 0)}^{2} - 2$

5 {(x + 0)}^{2} - 2

$5 {(x + 0)}^{2} - 2$

Step 1.2

Set

y

$y$ equal to the new right side.

y = 5 {(x + 0)}^{2} - 2

$y = 5 {(x + 0)}^{2} - 2$

y = 5 {(x + 0)}^{2} - 2

$y = 5 {(x + 0)}^{2} - 2$

Step 2

Use the vertex form,

y = a {(x - h)}^{2} + k

$y = a {(x - h)}^{2} + k$ , to determine the values of

a

$a$ ,

h

$h$ , and

k

$k$ .

a = 5

$a = 5$

h = 0

$h = 0$

k = - 2

$k = - 2$

Step 3

Find the vertex

(h, k)

$(h, k)$ .

(0, - 2)

$(0, - 2)$

Step 4

Enter YOUR Problem

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

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