Find the vertex and axis of symmetry and determine the direction that the parabola opens.
$x=-2y^2+6$

• A. Parabola opens to the left; Vertex: $(0,6)$; Axis of Symmetry: $y=0$
• B. Parabola opens to the right; Vertex: $\left(6,0\right)$; Axis of Symmetry: $y=2$
• C. Parabola opens to the left; Vertex: $(6,0)$; Axis of Symmetry: $y=0$
• D. Parabola opens to the right; Vertex: $\left(0,6\right)$; Axis of Symmetry: $y=2$

Answer: (C)

Verified step by step guidance

1. Rewrite the given equation \(x = -2y^2 + 6\) in the form \(x = a(y - k)^2 + h\) to identify the vertex \((h, k)\). Here, the equation is already in vertex form with \(a = -2\), \(h = 6\), and \(k = 0\), so the vertex is \((6, 0)\).
2. Determine the axis of symmetry. Since the equation is expressed as \(x\) in terms of \(y\), the axis of symmetry is a horizontal line given by \(y = k\). From the vertex, \(k = 0\), so the axis of symmetry is \(y = 0\).
3. Analyze the coefficient \(a = -2\) to determine the direction the parabola opens. Because \(a\) is negative and the parabola is in the form \(x = a(y - k)^2 + h\), the parabola opens to the left (toward decreasing \(x\) values).
4. Summarize the findings: the vertex is at \((6, 0)\), the axis of symmetry is the line \(y = 0\), and the parabola opens to the left.
5. If needed, sketch the parabola using the vertex and axis of symmetry as references, noting the direction it opens based on the sign of \(a\).