How do you graph the parabola #y = (x-2)^2 +2# using vertex, intercepts and additional points?

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Answer 1 · 2018-02-07T18:13:37

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#y-2=(x-2)^2#
It is of the form #Y=4AX^2# [an upward parabola]

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Vertex is (2,2)

*For y intercept put x=0 we get y=6 *

Some points (0,6),(2,2) (4,6)

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Final Graph
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