Precalculus Examples

x^{2} + y^{2} + 1 + 2 x - y = 0

$x^{2} + y^{2} + 1 + 2 x - y = 0$

Step 1

Subtract

1

$1$ from both sides of the equation.

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

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

Step 2

Complete the square for

x^{2} + 2 x

$x^{2} + 2 x$ .

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Step 2.1

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 = 1

$a = 1$

b = 2

$b = 2$

c = 0

$c = 0$

Step 2.2

Consider the vertex form of a parabola.

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

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

Step 2.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

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

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Step 2.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

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

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

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

Step 2.3.2

Cancel the common factor of

2

$2$ .

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Step 2.3.2.1

Cancel the common factor.

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

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

Step 2.3.2.2

Rewrite the expression.

d = 1

$d = 1$

d = 1

$d = 1$

d = 1

$d = 1$

Step 2.4

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 2.4.1

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

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

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Raise

2

$2$ to the power of

2

$2$ .

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

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

Step 2.4.2.1.2

Multiply

4

$4$ by

1

$1$ .

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

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

Step 2.4.2.1.3

Cancel the common factor of

4

$4$ .

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Step 2.4.2.1.3.1

Cancel the common factor.

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

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

Step 2.4.2.1.3.2

Rewrite the expression.

e = 0 - 1 \cdot 1

$e = 0 - 1 \cdot 1$

e = 0 - 1 \cdot 1

$e = 0 - 1 \cdot 1$

Step 2.4.2.1.4

Multiply

- 1

$- 1$ by

1

$1$ .

e = 0 - 1

$e = 0 - 1$

e = 0 - 1

$e = 0 - 1$

Step 2.4.2.2

Subtract

1

$1$ from

0

$0$ .

e = - 1

$e = - 1$

e = - 1

$e = - 1$

e = - 1

$e = - 1$

Step 2.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

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

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

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

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

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

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

Step 3

Substitute

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

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

x^{2} + 2 x

$x^{2} + 2 x$ in the equation

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

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

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

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

Step 4

Move

- 1

$- 1$ to the right side of the equation by adding

1

$1$ to both sides.

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

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

Step 5

Complete the square for

y^{2} - y

$y^{2} - y$ .

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Step 5.1

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 = 1

$a = 1$

b = - 1

$b = - 1$

c = 0

$c = 0$

Step 5.2

Consider the vertex form of a parabola.

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

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

Step 5.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

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

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Step 5.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

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

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

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

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Cancel the common factor of

- 1

$- 1$ and

1

$1$ .

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Step 5.3.2.1.1

Rewrite

- 1

$- 1$ as

- 1 (1)

$- 1 (1)$ .

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

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

Step 5.3.2.1.2

Cancel the common factor.

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

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

Step 5.3.2.1.3

Rewrite the expression.

d = \frac{- 1}{2}

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

d = \frac{- 1}{2}

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

Step 5.3.2.2

Move the negative in front of the fraction.

d = - \frac{1}{2}

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

d = - \frac{1}{2}

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

d = - \frac{1}{2}

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

Step 5.4

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 5.4.1

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

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

Step 5.4.2

Simplify the right side.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Raise

- 1

$- 1$ to the power of

2

$2$ .

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

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

Step 5.4.2.1.2

Multiply

4

$4$ by

1

$1$ .

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

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

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

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

Step 5.4.2.2

Subtract

\frac{1}{4}

$\frac{1}{4}$ from

0

$0$ .

e = - \frac{1}{4}

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

e = - \frac{1}{4}

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

e = - \frac{1}{4}

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

Step 5.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

{(y - \frac{1}{2})}^{2} - \frac{1}{4}

${(y - \frac{1}{2})}^{2} - \frac{1}{4}$ .

{(y - \frac{1}{2})}^{2} - \frac{1}{4}

${(y - \frac{1}{2})}^{2} - \frac{1}{4}$

{(y - \frac{1}{2})}^{2} - \frac{1}{4}

${(y - \frac{1}{2})}^{2} - \frac{1}{4}$

Step 6

Substitute

{(y - \frac{1}{2})}^{2} - \frac{1}{4}

${(y - \frac{1}{2})}^{2} - \frac{1}{4}$ for

y^{2} - y

$y^{2} - y$ in the equation

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

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

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

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

Step 7

Move

- \frac{1}{4}

$- \frac{1}{4}$ to the right side of the equation by adding

\frac{1}{4}

$\frac{1}{4}$ to both sides.

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

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

Step 8

Simplify

- 1 + 1 + \frac{1}{4}

$- 1 + 1 + \frac{1}{4}$ .

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Step 8.1

Find the common denominator.

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Step 8.1.1

Write

- 1

$- 1$ as a fraction with denominator

1

$1$ .

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

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

Step 8.1.2

Multiply

\frac{- 1}{1}

$\frac{- 1}{1}$ by

\frac{4}{4}

$\frac{4}{4}$ .

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1}{1} \cdot \frac{4}{4} + 1 + \frac{1}{4}

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

Step 8.1.3

Multiply

\frac{- 1}{1}

$\frac{- 1}{1}$ by

\frac{4}{4}

$\frac{4}{4}$ .

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1 \cdot 4}{4} + 1 + \frac{1}{4}

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

Step 8.1.4

Write

1

$1$ as a fraction with denominator

1

$1$ .

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1 \cdot 4}{4} + \frac{1}{1} + \frac{1}{4}

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

Step 8.1.5

Multiply

\frac{1}{1}

$\frac{1}{1}$ by

\frac{4}{4}

$\frac{4}{4}$ .

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1 \cdot 4}{4} + \frac{1}{1} \cdot \frac{4}{4} + \frac{1}{4}

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

Step 8.1.6

Multiply

\frac{1}{1}

$\frac{1}{1}$ by

\frac{4}{4}

$\frac{4}{4}$ .

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1 \cdot 4}{4} + \frac{4}{4} + \frac{1}{4}

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

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1 \cdot 4}{4} + \frac{4}{4} + \frac{1}{4}

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

Step 8.2

Combine the numerators over the common denominator.

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{- 1 \cdot 4 + 4 + 1}{4}

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

Step 8.3

Simplify the expression.

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Step 8.3.1

Multiply

- 1

$- 1$ by

4

$4$ .

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

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

Step 8.3.2

Add

- 4

$- 4$ and

4

$4$ .

{(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{0 + 1}{4}

${(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{0 + 1}{4}$

Step 8.3.3

Add

0

$0$ and

1

$1$ .

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

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

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

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

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

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

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