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: $(-2,-2)$; 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: $\(\left\)(-2,-2\(\right\))$; 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. Identify the given equation of the parabola: \(x = y^2 + 4y + 2\). Notice that the parabola is expressed with \(x\) as a function of \(y\), which means it opens either to the left or to the right, not up or down.
2. To find the vertex, rewrite the equation in vertex form by completing the square for the \(y\) terms. Start by grouping the \(y\) terms: \(x = (y^2 + 4y) + 2\).
3. Complete the square inside the parentheses: 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 parentheses to keep the equation balanced: \(x = (y^2 + 4y + 4 - 4) + 2\).
4. Rewrite the perfect square trinomial as a binomial squared and simplify constants: \(x = (y + 2)^2 - 4 + 2\), which simplifies to \(x = (y + 2)^2 - 2\). This is the vertex form of the parabola.
5. From the vertex form \(x = (y + 2)^2 - 2\), identify the vertex as \((-2, -2)\) (since the vertex is at \((x, y)\) corresponding to the values that make the squared term zero). The axis of symmetry is the line \(y = -2\). Because the squared term is positive, the parabola opens to the right.