Answer:
A. y = -2
B. A(6, -2)
C. use the midpoint tool to locate the vertex B(6, 1) halfway between F and A
D. up
E. divide 6 by 2; positive
F. 3
G. y = (1/12)(x -6)² +1
H. see attached
Step-by-step explanation:
A. The horizontal line 6 units below a point with y-coordinate 4 will be at ...
y = 4 -6
y = -2
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B. The point of intersection will have the same x-coordinate as the focus (6) and the same y-coordinate as the line (-2). That point is A(6, -2).
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C. As with every other point on the parabola, the vertex is the same distance from the focus as it is from the directrix. It will be the midpoint of segment FA. The vertex is B(6, 1).
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D. The focus is "inside" the parabola. The focus is above the directrix, so the parabola opens upward, away from the directrix.
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E, F. We're told to locate the directrix 6 units below the focus. The value of p is half that distance, 3 units. It can also be found by using GeoGebra to measure the length of segment FB.
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G. Using the vertex form equation with p=3, (h, k) = (6, 1), the equation of the parabola is ...

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H. see the attachment
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I. Multiplying the above equation by 12 and eliminating parentheses, we have ...
12y = x^2 -12x +36 +12
-48 = x^2 -12x -12y . . . . . . . subtract 48+12y
This is the same equation as shown by GeoGebra.