Precalculus Examples

$x^{2} + y^{2} + 2 x + 2 y = 1$

Step 1

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

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Step 1.1

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

$a = 1$

$b = 2$

$c = 0$

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

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

Step 1.3.2

Cancel the common factor of $2$ .

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

$d = 1$

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

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

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 $2$ to the power of $2$ .

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

Step 1.4.2.1.2

Multiply $4$ by $1$ .

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

Step 1.4.2.1.3

Cancel the common factor of $4$ .

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Step 1.4.2.1.3.1

Cancel the common factor.

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

Step 1.4.2.1.3.2

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Step 1.4.2.1.4

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Step 1.4.2.2

Subtract $1$ from $0$ .

$e = - 1$

Step 1.5

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

${(x + 1)}^{2} - 1$

Step 2

Substitute ${(x + 1)}^{2} - 1$ for $x^{2} + 2 x$ in the equation $x^{2} + y^{2} + 2 x + 2 y = 1$ .

${(x + 1)}^{2} - 1 + y^{2} + 2 y = 1$

Step 3

Move $- 1$ to the right side of the equation by adding $1$ to both sides.

${(x + 1)}^{2} + y^{2} + 2 y = 1 + 1$

Step 4

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

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Step 4.1

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

$a = 1$

$b = 2$

$c = 0$

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

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

Step 4.3.2

Cancel the common factor of $2$ .

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Step 4.3.2.1

Cancel the common factor.

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

Step 4.3.2.2

Rewrite the expression.

$d = 1$

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

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

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 $2$ to the power of $2$ .

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

Step 4.4.2.1.2

Multiply $4$ by $1$ .

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

Step 4.4.2.1.3

Cancel the common factor of $4$ .

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Step 4.4.2.1.3.1

Cancel the common factor.

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

Step 4.4.2.1.3.2

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Step 4.4.2.1.4

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Step 4.4.2.2

Subtract $1$ from $0$ .

$e = - 1$

Step 4.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 1)}^{2} - 1$ .

${(y + 1)}^{2} - 1$

Step 5

Substitute ${(y + 1)}^{2} - 1$ for $y^{2} + 2 y$ in the equation $x^{2} + y^{2} + 2 x + 2 y = 1$ .

${(x + 1)}^{2} + {(y + 1)}^{2} - 1 = 1 + 1$

Step 6

Move $- 1$ to the right side of the equation by adding $1$ to both sides.

${(x + 1)}^{2} + {(y + 1)}^{2} = 1 + 1 + 1$

Step 7

Simplify $1 + 1 + 1$ .

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Step 7.1

Add $1$ and $1$ .

${(x + 1)}^{2} + {(y + 1)}^{2} = 2 + 1$

Step 7.2

Add $2$ and $1$ .

${(x + 1)}^{2} + {(y + 1)}^{2} = 3$

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