Precalculus Examples

$4 x^{2} + 4 y^{2} - 24 x + 32 y + 36 = 0$

Step 1

Subtract $36$ from both sides of the equation.

$4 x^{2} + 4 y^{2} - 24 x + 32 y = - 36$

Step 2

Divide both sides of the equation by $4$ .

$x^{2} + y^{2} - 6 x + 8 y = - 9$

Step 3

Complete the square for $x^{2} - 6 x$ .

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Step 3.1

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

$a = 1$

$b = - 6$

$c = 0$

Step 3.2

Consider the vertex form of a parabola.

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Cancel the common factor of $- 6$ and $2$ .

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Step 3.3.2.1

Factor $2$ out of $- 6$ .

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

Step 3.3.2.2

Cancel the common factors.

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Step 3.3.2.2.1

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

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

Step 3.3.2.2.2

Cancel the common factor.

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

Step 3.3.2.2.3

Rewrite the expression.

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

Step 3.3.2.2.4

Divide $- 3$ by $1$ .

$d = - 3$

Step 3.4

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

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Step 3.4.1

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

$e = 0 - \frac{{(- 6)}^{2}}{4 \cdot 1}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Raise $- 6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot 1}$

Step 3.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{36}{4}$

Step 3.4.2.1.3

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Step 3.4.2.1.4

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Step 3.4.2.2

Subtract $9$ from $0$ .

$e = - 9$

Step 3.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 3)}^{2} - 9$ .

${(x - 3)}^{2} - 9$

Step 4

Substitute ${(x - 3)}^{2} - 9$ for $x^{2} - 6 x$ in the equation $x^{2} + y^{2} - 6 x + 8 y = - 9$ .

${(x - 3)}^{2} - 9 + y^{2} + 8 y = - 9$

Step 5

Move $- 9$ to the right side of the equation by adding $9$ to both sides.

${(x - 3)}^{2} + y^{2} + 8 y = - 9 + 9$

Step 6

Complete the square for $y^{2} + 8 y$ .

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Step 6.1

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

$a = 1$

$b = 8$

$c = 0$

Step 6.2

Consider the vertex form of a parabola.

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

Step 6.3

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

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Step 6.3.1

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

$d = \frac{8}{2 \cdot 1}$

Step 6.3.2

Cancel the common factor of $8$ and $2$ .

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Step 6.3.2.1

Factor $2$ out of $8$ .

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Step 6.3.2.2

Cancel the common factors.

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Step 6.3.2.2.1

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

$d = \frac{2 \cdot 4}{2 (1)}$

Step 6.3.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Step 6.3.2.2.3

Rewrite the expression.

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

Step 6.3.2.2.4

Divide $4$ by $1$ .

$d = 4$

Step 6.4

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

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Step 6.4.1

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

$e = 0 - \frac{8^{2}}{4 \cdot 1}$

Step 6.4.2

Simplify the right side.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

Raise $8$ to the power of $2$ .

$e = 0 - \frac{64}{4 \cdot 1}$

Step 6.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{64}{4}$

Step 6.4.2.1.3

Divide $64$ by $4$ .

$e = 0 - 1 \cdot 16$

Step 6.4.2.1.4

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Step 6.4.2.2

Subtract $16$ from $0$ .

$e = - 16$

Step 6.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 4)}^{2} - 16$ .

${(y + 4)}^{2} - 16$

Step 7

Substitute ${(y + 4)}^{2} - 16$ for $y^{2} + 8 y$ in the equation $x^{2} + y^{2} - 6 x + 8 y = - 9$ .

${(x - 3)}^{2} + {(y + 4)}^{2} - 16 = - 9 + 9$

Step 8

Move $- 16$ to the right side of the equation by adding $16$ to both sides.

${(x - 3)}^{2} + {(y + 4)}^{2} = - 9 + 9 + 16$

Step 9

Simplify $- 9 + 9 + 16$ .

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Step 9.1

Add $- 9$ and $9$ .

${(x - 3)}^{2} + {(y + 4)}^{2} = 0 + 16$

Step 9.2

Add $0$ and $16$ .

${(x - 3)}^{2} + {(y + 4)}^{2} = 16$