College Algebra

The graph of a quadratic function f(x) is shown above. It has a vertex at (-2,4) and passes the point (0,2). Find the quadratic function

y=a(x-(-2))^2+4
y=a(x+2)+4
2=a(x+2)+4
-4 -4
-2=a(0+2)/2
-1+ a

function -1(x+2)+4 ?
Im not sure where Im going wrong but my answer was incorrect. Please help

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asked by Anonymous
  1. from the given vertex, the equation must be

    y = a(x+2)^2 + 4
    but (0,2) lies on it, so
    2 = a(0+2)^2 + 4
    2 = 4a + 4
    4a=-2
    a = -1/2

    y = (-1/2)(x+2)^2 + 4

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    posted by Reiny

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