Find the vertex of y^2 + 4y + 2x +10 = 0. Use completing the square to put the equation into standard form as your first step. Could you show me how to get the vertex? Thanks

y^2 + 4y + 4 - 4 + 2x + 10 = 0

(y+2)^2 + 6 = -2x
x = (-1/2)(y+2)^2 + 6

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To find the vertex of the quadratic equation, we first need to rewrite the equation in standard form by completing the square. The standard form of a quadratic equation is given by:

y = a(x-h)^2 + k,

where (h, k) represents the coordinates of the vertex.

Let's complete the square for the given equation, y^2 + 4y + 2x + 10 = 0:

1. Move the constant term (10) to the other side of the equation:
y^2 + 4y + 2x = -10

2. Group the terms with y together:
(y^2 + 4y) + 2x = -10

3. Take half of the coefficient of y (4/2 = 2), square it (2^2 = 4), and add it to both sides:
(y^2 + 4y + 4) + 2x = -10 + 4

4. Rewrite the expression inside the parentheses as a perfect square:
(y + 2)^2 + 2x = -6

Now the equation is in standard form. We can see that the vertex is given by the coordinates (-2, -6). The value of h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.