Determine the Vertex and Axis of Symmetry for the parabola $x=y^2+4y+2$, and determine which direction the parabola will open.

• A. The parabola opens upwards; Vertex: $\(\left\)(-2,-2\(\right\))$; Axis of Symmetry: $y=-2$
• B. The parabola opens to the right; Vertex: $(2,2)$; Axis of Symmetry: $y=-2$
• C. The parabola opens to the right; Vertex: $(-2,-2)$; Axis of Symmetry: $y=-2$
• D. The parabola opens upwards; Vertex: $\(\left\)(-2,-2\(\right\))$; Axis of Symmetry: $x=-2$

Answer: (C)

Verified step by step guidance

1. Rewrite the given equation \(x = y^2 + 4y + 2\) to identify it as a parabola that opens horizontally (since \(x\) is expressed in terms of \(y\)).
2. Complete the square for the \(y\)-terms on the right side to rewrite the equation in vertex form. Start by grouping the \(y\) terms: \(y^2 + 4y\).
3. Find the value to complete the square: take half of the coefficient of \(y\) (which is 4), divide by 2 to get 2, then square it to get 4. Add and subtract 4 inside the equation to maintain equality.
4. Rewrite the equation as \(x = (y + 2)^2 + (2 - 4)\), simplifying the constant terms to get the vertex form \(x = (y + 2)^2 - 2\).
5. From the vertex form, identify the vertex as \((-2, -2)\) (since the equation is \(x = (y - (-2))^2 - 2\)), the axis of symmetry as the horizontal line \(y = -2\), and determine that the parabola opens to the right because the squared term is positive and \(x\) is expressed in terms of \(y\).