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: $(6,0)$; Axis of Symmetry: $y=2$
• C. Parabola opens to the left; Vertex: $\left(6,0\right)$; 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 standard form of a parabola that opens horizontally, which is \(x = a(y - k)^2 + h\), where \((h, k)\) is the vertex.
2. Identify the vertex \((h, k)\) by comparing the given equation to the standard form. Here, \(h\) is the constant term and \(k\) is the value that makes the squared term zero. So, the vertex is at \((6, 0)\).
3. Determine the axis of symmetry, which is the line that passes through the vertex and is perpendicular to the direction the parabola opens. Since the parabola opens horizontally, the axis of symmetry is a horizontal line \(y = k\), so here it is \(y = 0\).
4. Analyze the coefficient \(a = -2\) in front of the squared term. Because \(a\) is negative, the parabola opens in the negative \(x\)-direction, which means it opens to the left.
5. 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.