Precalculus Examples

$2 x^{2} + 2 y^{2} + 8 x + 7 = 0$

Step 1

Subtract $7$ from both sides of the equation.

$2 x^{2} + 2 y^{2} + 8 x = - 7$

Step 2

Divide both sides of the equation by $2$ .

$x^{2} + y^{2} + 4 x = - \frac{7}{2}$

Step 3

Complete the square for $x^{2} + 4 x$ .

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Step 3.1

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

$a = 1$

$b = 4$

$c = 0$

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

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

Step 3.3.2

Cancel the common factor of $4$ and $2$ .

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Step 3.3.2.1

Factor $2$ out of $4$ .

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

Step 3.3.2.2

Cancel the common factors.

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Step 3.3.2.2.1

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

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

Step 3.3.2.2.2

Cancel the common factor.

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

Step 3.3.2.2.3

Rewrite the expression.

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

Step 3.3.2.2.4

Divide $2$ by $1$ .

$d = 2$

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

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

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

Cancel the common factor of $4^{2}$ and $4$ .

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Step 3.4.2.1.1.1

Factor $4$ out of $4^{2}$ .

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

Step 3.4.2.1.1.2

Cancel the common factors.

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Step 3.4.2.1.1.2.1

Factor $4$ out of $4 \cdot 1$ .

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

Step 3.4.2.1.1.2.2

Cancel the common factor.

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

Step 3.4.2.1.1.2.3

Rewrite the expression.

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

Step 3.4.2.1.1.2.4

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Step 3.4.2.1.2

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Step 3.4.2.2

Subtract $4$ from $0$ .

$e = - 4$

Step 3.5

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

${(x + 2)}^{2} - 4$

Step 4

Substitute ${(x + 2)}^{2} - 4$ for $x^{2} + 4 x$ in the equation $x^{2} + y^{2} + 4 x = - \frac{7}{2}$ .

${(x + 2)}^{2} - 4 + y^{2} = - \frac{7}{2}$

Step 5

Move $- 4$ to the right side of the equation by adding $4$ to both sides.

${(x + 2)}^{2} + y^{2} = - \frac{7}{2} + 4$

Step 6

Simplify $- \frac{7}{2} + 4$ .

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Step 6.1

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

${(x + 2)}^{2} + y^{2} = - \frac{7}{2} + 4 \cdot \frac{2}{2}$

Step 6.2

Combine $4$ and $\frac{2}{2}$ .

${(x + 2)}^{2} + y^{2} = - \frac{7}{2} + \frac{4 \cdot 2}{2}$

Step 6.3

Combine the numerators over the common denominator.

${(x + 2)}^{2} + y^{2} = \frac{- 7 + 4 \cdot 2}{2}$

Step 6.4

Simplify the numerator.

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Step 6.4.1

Multiply $4$ by $2$ .

${(x + 2)}^{2} + y^{2} = \frac{- 7 + 8}{2}$

Step 6.4.2

Add $- 7$ and $8$ .

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

Step 7

This is the form of a circle. Use this form to determine the center and radius of the circle.

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Step 8

Match the values in this circle to those of the standard form. The variable $r$ represents the radius of the circle, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from origin.

$r = \frac{\sqrt{2}}{2}$

$h = - 2$

$k = 0$

Step 9

The center of the circle is found at $(h, k)$ .

Center: $(- 2, 0)$

Step 10

These values represent the important values for graphing and analyzing a circle.

Center: $(- 2, 0)$

Radius: $\frac{\sqrt{2}}{2}$

Step 11