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The table of values represents points on the graph of a quadratic function.

x -2 -1 0 1 2
y 31 16 7 4 7

I got Vertex (1,4) & f(x) = -7x^2 + 14x - 3 as my equation (standard form).
May someone check my work please

3 answers

I like the vertex, but clearly the graph opens upward, so you cannot have -7x^2

show your work, and I think you will spot your own mistake. Start with
y = a(x-1)^2 + 4
and just find a
On second try I got y= 3x^2 -6x + 7

Does that look good?
looks better. I assume you checked to make sure the points fit.
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