In Exercises 1-4, write the standard form of the quadratic function whose graph is a parabola with
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1. Vertex: (−2, 5); point: (0, 9)
2. Vertex: (−3, −10); point: (0, 8)
3. Vertex: (1, −2); point: (−1, 14)
4. Vertex: (2, 3); point: (0, 2)
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1 2 5 is the vertex fx ax 2 2 5 Because the graph passes through 09 9 a0 2 2 5 4 4a 1 a So fx 1x 2 ...View the full answer
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