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1.
Find the vertex, focus, directrix, and focal width of the parabola.

x2 = 12y (1 point)
Vertex: (0, 0); Focus: (3, 0); Directrix: x = 3; Focal width: 3
Vertex: (0, 0); Focus: (0, 3); Directrix: y = -3; Focal width: 12
Vertex: (0, 0); Focus: (3, 0); Directrix: y = 3; Focal width: 48
Vertex: (0, 0); Focus: (0, -3); Directrix: x = -3; Focal width: 48
2.
Find the vertex, focus, directrix, and focal width of the parabola.

x = 3y2 (2 points)
Vertex: (0, 0); Focus: ; Directrix: x = ; Focal width: 12
Vertex: (0, 0); Focus: ; Directrix: x = ; Focal width: 0.33
Vertex: (0, 0); Focus: ; Directrix: x = ; Focal width: 0.33
Vertex: (0, 0); Focus: ; Directrix: y = ; Focal width: 12
3.
Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7). (1 point)
y = x2
y2 = -7x
y = x2
y2 = -28x
4.
Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8. (2 points)
y = x2
y2 = 8x
y2 = 32x
y = x2
5.
Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7. (2 points)
y = x2
x = y2
-28y = x2
y2 = 14x
6.
A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base, as shown below.Find an equation for the parabola if the vertex is put at the origin of the coordinate system. (2 points)

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