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xoxomath
Jun 4, 2012, 07:24 PM
find the vertex of y=2x^2-8x+3

ebaines
Jun 5, 2012, 05:57 AM
find the vertex of y=2x^2-8x+3

You neeed to rewrite this equation in the standard form for parabolas, which is:


y-y_0 = \frac 1 {4p} (x-x_0)^2


Once in this form the vertex is given by (x_0, y_0).