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

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

Answer: (C)

Verified step by step guidance

1. Identify the given equation of the parabola: \(x = 6y^2\). Notice that the variable \(x\) is expressed in terms of \(y^2\), which means this is a parabola that opens horizontally (either to the right or left), not vertically.
2. Recall the standard form of a horizontally opening parabola: \(x = a(y - k)^2 + h\), where \((h, k)\) is the vertex. Compare the given equation to this form to find the vertex.
3. Since the equation is \(x = 6y^2\), it can be rewritten as \(x = 6(y - 0)^2 + 0\), so the vertex is at \((0, 0)\).
4. The axis of symmetry for a horizontally opening parabola is a horizontal line through the vertex, which means it is \(y = k\). Here, since \(k = 0\), the axis of symmetry is \(y = 0\).
5. Determine the direction the parabola opens by looking at the coefficient \(a = 6\). Because \(a\) is positive, the parabola opens to the right. If \(a\) were negative, it would open to the left.