Algebra
write the equation in standard form and then graph the equation.
y=2x^28x+11

You will have to use "completing the square".
y = 2x^28x+11
= 2(x^2  4x+ .... ) + 11 < took out the coefficient of the square term from the first two terms
= 2(x^2  4x + 4  4) < took half of the coefficient of the x term, then squared it, added it and immediately subtracted it again, the net result was adding zero.
= 2( (x2)^2  4) + 11 < formed the perfect square plus the left over constant
= 2(x2)^2  8 + 11 < dististributed the 2 over the perfect square and the leftover constant
= 2(x2)^2 + 3 < added the stuff at the end
all done, now I can read the vertex
which is .......
and if I let x = 0 in the original (or this last) equation, I get
y = 11
So plot the vertex, plot the yintercept and form a neat parabola from the vertex to hit (0,11). Complete the other half of the parabola.
It should look like this:
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E28x%2B11,+y+%3D+2(x2)%5E2+%2B+3
notice I graphed both forms of the equation and the graphs coincide. This was a neat way to check my work.posted by Reiny
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