Algebra

write the equation in standard form and then graph the equation.
y=2x^2-8x+11

asked by Anonymous
  1. You will have to use "completing the square".

    y = 2x^2-8x+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( (x-2)^2 - 4) + 11 <----- formed the perfect square plus the left over constant
    = 2(x-2)^2 - 8 + 11 <---- dististributed the 2 over the perfect square and the left-over constant
    = 2(x-2)^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 y-intercept 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%5E2-8x%2B11,+y+%3D+2(x-2)%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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