Precalculus Examples

$x^{2} + 12 x + 3 y^{2} - 12 y = - 3$

Step 1

Complete the square for $x^{2} + 12 x$ .

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Step 1.1

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

$a = 1$

$b = 12$

$c = 0$

Step 1.2

Consider the vertex form of a parabola.

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

Step 1.3

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

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Step 1.3.1

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

$d = \frac{12}{2 \cdot 1}$

Step 1.3.2

Cancel the common factor of $12$ and $2$ .

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Step 1.3.2.1

Factor $2$ out of $12$ .

$d = \frac{2 \cdot 6}{2 \cdot 1}$

Step 1.3.2.2

Cancel the common factors.

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Step 1.3.2.2.1

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 6}{2 (1)}$

Step 1.3.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot 6}{2 \cdot 1}$

Step 1.3.2.2.3

Rewrite the expression.

$d = \frac{6}{1}$

Step 1.3.2.2.4

Divide $6$ by $1$ .

$d = 6$

Step 1.4

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

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Step 1.4.1

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

$e = 0 - \frac{12^{2}}{4 \cdot 1}$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Raise $12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot 1}$

Step 1.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{144}{4}$

Step 1.4.2.1.3

Divide $144$ by $4$ .

$e = 0 - 1 \cdot 36$

Step 1.4.2.1.4

Multiply $- 1$ by $36$ .

$e = 0 - 36$

Step 1.4.2.2

Subtract $36$ from $0$ .

$e = - 36$

Step 1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 6)}^{2} - 36$ .

${(x + 6)}^{2} - 36$

Step 2

Substitute ${(x + 6)}^{2} - 36$ for $x^{2} + 12 x$ in the equation $x^{2} + 12 x + 3 y^{2} - 12 y = - 3$ .

${(x + 6)}^{2} - 36 + 3 y^{2} - 12 y = - 3$

Step 3

Move $- 36$ to the right side of the equation by adding $36$ to both sides.

${(x + 6)}^{2} + 3 y^{2} - 12 y = - 3 + 36$

Step 4

Complete the square for $3 y^{2} - 12 y$ .

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Step 4.1

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

$a = 3$

$b = - 12$

$c = 0$

Step 4.2

Consider the vertex form of a parabola.

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

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

$d = \frac{- 12}{2 \cdot 3}$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Cancel the common factor of $- 12$ and $2$ .

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Step 4.3.2.1.1

Factor $2$ out of $- 12$ .

$d = \frac{2 \cdot - 6}{2 \cdot 3}$

Step 4.3.2.1.2

Cancel the common factors.

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Step 4.3.2.1.2.1

Factor $2$ out of $2 \cdot 3$ .

$d = \frac{2 \cdot - 6}{2 (3)}$

Step 4.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 6}{2 \cdot 3}$

Step 4.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 6}{3}$

Step 4.3.2.2

Cancel the common factor of $- 6$ and $3$ .

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Step 4.3.2.2.1

Factor $3$ out of $- 6$ .

$d = \frac{3 \cdot - 2}{3}$

Step 4.3.2.2.2

Cancel the common factors.

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Step 4.3.2.2.2.1

Factor $3$ out of $3$ .

$d = \frac{3 \cdot - 2}{3 (1)}$

Step 4.3.2.2.2.2

Cancel the common factor.

$d = \frac{3 \cdot - 2}{3 \cdot 1}$

Step 4.3.2.2.2.3

Rewrite the expression.

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

Step 4.3.2.2.2.4

Divide $- 2$ by $1$ .

$d = - 2$

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

$e = 0 - \frac{{(- 12)}^{2}}{4 \cdot 3}$

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

Raise $- 12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot 3}$

Step 4.4.2.1.2

Multiply $4$ by $3$ .

$e = 0 - \frac{144}{12}$

Step 4.4.2.1.3

Divide $144$ by $12$ .

$e = 0 - 1 \cdot 12$

Step 4.4.2.1.4

Multiply $- 1$ by $12$ .

$e = 0 - 12$

Step 4.4.2.2

Subtract $12$ from $0$ .

$e = - 12$

Step 4.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(y - 2)}^{2} - 12$ .

$3 {(y - 2)}^{2} - 12$

Step 5

Substitute $3 {(y - 2)}^{2} - 12$ for $3 y^{2} - 12 y$ in the equation $x^{2} + 12 x + 3 y^{2} - 12 y = - 3$ .

${(x + 6)}^{2} + 3 {(y - 2)}^{2} - 12 = - 3 + 36$

Step 6

Move $- 12$ to the right side of the equation by adding $12$ to both sides.

${(x + 6)}^{2} + 3 {(y - 2)}^{2} = - 3 + 36 + 12$

Step 7

Simplify $- 3 + 36 + 12$ .

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Step 7.1

Add $- 3$ and $36$ .

${(x + 6)}^{2} + 3 {(y - 2)}^{2} = 33 + 12$

Step 7.2

Add $33$ and $12$ .

${(x + 6)}^{2} + 3 {(y - 2)}^{2} = 45$

Step 8

Divide each term by $45$ to make the right side equal to one.

$\frac{{(x + 6)}^{2}}{45} + \frac{3 {(y - 2)}^{2}}{45} = \frac{45}{45}$

Step 9

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x + 6)}^{2}}{45} + \frac{{(y - 2)}^{2}}{15} = 1$