Precalculus Examples

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

Step 1

Write $f (x) = 2 x^{2} + x - 6$ as an equation.

$y = 2 x^{2} + x - 6$

Step 2

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 x^{2} + x - 6 = y$

Step 3

Add $6$ to both sides of the equation.

$2 x^{2} + x = y + 6$

Step 4

Complete the square.

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Step 4.1

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

$a = 2$

$b = 1$

$c = 0 = y + 6$

Step 4.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = y + 6$

Step 4.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 4.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{1}{2 \cdot 2}$

Step 4.3.2

Multiply $2$ by $2$ .

$d = \frac{1}{4}$

Step 4.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 4.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{1^{2}}{4 \cdot 2}$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

One to any power is one.

$e = 0 - \frac{1}{4 \cdot 2}$

Step 4.4.2.1.2

Multiply $4$ by $2$ .

$e = 0 - \frac{1}{8}$

Step 4.4.2.2

Subtract $\frac{1}{8}$ from $0$ .

$e = - \frac{1}{8}$

Step 4.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}$ .

$2 {(x + \frac{1}{4})}^{2} - \frac{1}{8} = y + 6$

Step 5

Move all terms not containing $x$ to the right side of the equation.

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Step 5.1

Add $\frac{1}{8}$ to both sides of the equation.

$2 {(x + \frac{1}{4})}^{2} = y + 6 + \frac{1}{8}$

Step 5.2

To write $6$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$2 {(x + \frac{1}{4})}^{2} = y + 6 \cdot \frac{8}{8} + \frac{1}{8}$

Step 5.3

Combine $6$ and $\frac{8}{8}$ .

$2 {(x + \frac{1}{4})}^{2} = y + \frac{6 \cdot 8}{8} + \frac{1}{8}$

Step 5.4

Combine the numerators over the common denominator.

$2 {(x + \frac{1}{4})}^{2} = y + \frac{6 \cdot 8 + 1}{8}$

Step 5.5

Simplify the numerator.

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Step 5.5.1

Multiply $6$ by $8$ .

$2 {(x + \frac{1}{4})}^{2} = y + \frac{48 + 1}{8}$

Step 5.5.2

Add $48$ and $1$ .

$2 {(x + \frac{1}{4})}^{2} = y + \frac{49}{8}$

Step 6

Divide each term in $2 {(x + \frac{1}{4})}^{2} = y + \frac{49}{8}$ by $2$ and simplify.

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Step 6.1

Divide each term in $2 {(x + \frac{1}{4})}^{2} = y + \frac{49}{8}$ by $2$ .

$\frac{2 {(x + \frac{1}{4})}^{2}}{2} = \frac{y}{2} + \frac{\frac{49}{8}}{2}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{2 {(x + \frac{1}{4})}^{2}}{2} = \frac{y}{2} + \frac{\frac{49}{8}}{2}$

Step 6.2.1.2

Divide ${(x + \frac{1}{4})}^{2}$ by $1$ .

${(x + \frac{1}{4})}^{2} = \frac{y}{2} + \frac{\frac{49}{8}}{2}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Simplify each term.

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Step 6.3.1.1

Multiply the numerator by the reciprocal of the denominator.

${(x + \frac{1}{4})}^{2} = \frac{y}{2} + \frac{49}{8} \cdot \frac{1}{2}$

Step 6.3.1.2

Multiply $\frac{49}{8} \cdot \frac{1}{2}$ .

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Step 6.3.1.2.1

Multiply $\frac{49}{8}$ by $\frac{1}{2}$ .

${(x + \frac{1}{4})}^{2} = \frac{y}{2} + \frac{49}{8 \cdot 2}$

Step 6.3.1.2.2

Multiply $8$ by $2$ .

${(x + \frac{1}{4})}^{2} = \frac{y}{2} + \frac{49}{16}$

Step 7

Factor.

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Step 7.1

To write $\frac{y}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

${(x + \frac{1}{4})}^{2} = \frac{y}{2} \cdot \frac{8}{8} + \frac{49}{16}$

Step 7.2

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Step 7.2.1

Multiply $\frac{y}{2}$ by $\frac{8}{8}$ .

${(x + \frac{1}{4})}^{2} = \frac{y \cdot 8}{2 \cdot 8} + \frac{49}{16}$

Step 7.2.2

Multiply $2$ by $8$ .

${(x + \frac{1}{4})}^{2} = \frac{y \cdot 8}{16} + \frac{49}{16}$

Step 7.3

Combine the numerators over the common denominator.

${(x + \frac{1}{4})}^{2} = \frac{y \cdot 8 + 49}{16}$

Step 7.4

Move $8$ to the left of $y$ .

${(x + \frac{1}{4})}^{2} = \frac{8 y + 49}{16}$

Step 7.5

Reorder terms.

${(x + \frac{1}{4})}^{2} = \frac{1}{16} (8 y + 49)$