Find the Vertex Form x^2+4x+2y+y^2=9

$x^{2} + 4 x + 2 y + y^{2} = 9$

Step 1

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

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

$a = 1$

$b = 4$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

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

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

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2$ by $1$ .

$d = 2$

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

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

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

Simplify the right side.

Simplify each term.

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

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

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

Step 2

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

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

Step 3

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

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

Step 4

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

Reorder $2 y$ and $y^{2}$ .

$y^{2} + 2 y$

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

$a = 1$

$b = 2$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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 $2 y + y^{2}$ in the equation $x^{2} + 4 x + 2 y + y^{2} = 9$ .

${(x + 2)}^{2} + {(y + 1)}^{2} - 1 = 9 + 4$

Step 6

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

${(x + 2)}^{2} + {(y + 1)}^{2} = 9 + 4 + 1$

Step 7

Simplify $9 + 4 + 1$ .

Add $9$ and $4$ .

${(x + 2)}^{2} + {(y + 1)}^{2} = 13 + 1$

Add $13$ and $1$ .

${(x + 2)}^{2} + {(y + 1)}^{2} = 14$

Find the Vertex Form x^2+4x+2y+y^2=9

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