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what is the quadratic function that is created with roots at 2 and 4 and a vertex at (3 1)
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Well, we have the vertex form. y - k = a(x-h)^2 We are given the vertex (3,1) y - 1 = a(x-3)^2 We may wish to observe that 2-3 = -1 and 4-3 = 1. Thus, we have been given symmetric points, (2,0) and (4,0) and either should finish up for us. Substitute one of those points and solve for 'a'!
im still so confused on this one, i may be overthinking it
Substitute one of those points into the equation with 'a' remaining. Show me what you get,
y-1=2(x-3)^2?
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