triangle has vertices at A (1,3), B(4,2), and C (3,8). Which transformation would produce an image with vertices A'(3, -1), B(2,
2 answers:
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Answer:
see attached
Step-by-step explanation:
The equation is in the form ...
4p(y -k) = (x -h)^2 . . . . . (h, k) is the vertex; p is the focus-vertex distance
Comparing this to your equation, we see ...
p = 4, (h, k) = (3, 4)
p > 0, so the parabola opens upward. The vertex is on the axis of symmetry. That axis has the equation x=x-coordinate of vertex. This tells you ...
vertex: (3, 4)
axis of symmetry: x = 3
focus: (3, 8) . . . . . 4 units up from vertex
directrix: y = 0 . . . horizontal line 4 units down from vertex
