Find the Vertex x^2+4x+2y=0

$x^{2} + 4 x + 2 y = 0$

Step 1

Rewrite the equation in vertex form.

Isolate $y$ to the left side of the equation.

Move all terms not containing $y$ to the right side of the equation.

Subtract $x^{2}$ from both sides of the equation.

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

Subtract $4 x$ from both sides of the equation.

$2 y = - x^{2} - 4 x$

Divide each term in $2 y = - x^{2} - 4 x$ by $2$ and simplify.

Divide each term in $2 y = - x^{2} - 4 x$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

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

Factor $2$ out of $- 4 x$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 2 x$ by $1$ .

$y = - \frac{x^{2}}{2} - 2 x$

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

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

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

$b = - 2$

$c = 0$

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

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

Simplify the right side.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Dividing two negative values results in a positive value.

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

Multiply the numerator by the reciprocal of the denominator.

$d = 1 \cdot 2$

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

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

Simplify the right side.

Simplify each term.

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

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

Simplify the denominator.

Multiply $- 1$ by $4$ .

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

Combine $- 4$ and $\frac{1}{2}$ .

$e = 0 - \frac{4}{\frac{- 4}{2}}$

Divide $- 4$ by $2$ .

$e = 0 - \frac{4}{- 2}$

Divide $4$ by $- 2$ .

$e = 0 - - 2$

Multiply $- 1$ by $- 2$ .

$e = 0 + 2$

Add $0$ and $2$ .

$e = 2$

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

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

Set $y$ equal to the new right side.

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

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

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

$h = - 2$

$k = 2$

Step 3

Find the vertex $(h, k)$ .

$(- 2, 2)$

Step 4

Find the Vertex x^2+4x+2y=0

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