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Math Help - completing the square help

  1. #1
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    completing the square help

    2. By completing the square, sketch the curve
    y = 2x2 – 3x – 1
    Show the intercepts on the axes and
    the minimum point.
    Last edited by crashuk; October 24th 2008 at 11:12 AM.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by crashuk View Post
    completing the square,

    y=2x^2-3x-1,
    I assume you want to put this equation of a parabola in vertex form by completing the square.

     y=2x^2-3x-1

    First, factor out the 2 from the x terms:

    y=2\left(x^2-\frac{3}{2}x\right)-1

    Next, take half the coefficient of x, \left(-\frac{3}{4}\right), square it, \left(\frac{9}{16}\right), and add it to complete the perfect square trinomial in parentheses. But, you really added twice that amount since we have a factor of  2 outside the parentheses. So, to keep things balanced in the equation, we must subtract twice that amount, \left(\frac{9}{8}\right), outside the parentheses. 'What one addeth to just one side of an equation, one must taketh away'.

    y=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-1-\frac{9}{8}

    y=2\left(x-\frac{3}{4}\right)^2-\frac{17}{8}

    The minimum point is the y-coordinate of the vertex: -\frac{17}{8}

    The y-intercept can be found from the original function by setting x=0 and solving for y.

    The x-intercepts (zeros of the function) are found by setting the original quadratic = 0 and use the quadratic formula to find the values for x.
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  3. #3
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    Quote Originally Posted by masters View Post
    I assume you want to put this equation of a parabola in vertex form by completing the square.

     y=2x^2-3x-1

    First, factor out the 2 from the x terms:

    y=2\left(x^2-\frac{3}{2}x\right)-1

    Next, take half the coefficient of x, \left(-\frac{3}{4}\right), square it, \left(\frac{9}{16}\right), and add it to complete the perfect square trinomial in parentheses. But, you really added twice that amount since we have a factor of  2 outside the parentheses. So, to keep things balanced in the equation, we must subtract twice that amount, \left(\frac{9}{8}\right), outside the parentheses. 'What one addeth to just one side of an equation, one must taketh away'.

    y=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-1-\frac{9}{8}

    y=2\left(x-\frac{3}{4}\right)^2-\frac{17}{8}
    so i take it its the same for everything else if you want to complete the square.

    thanks for put some info, do you know any good web sites for someone just starting off in algebra
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by crashuk View Post
    so i take it its the same for everything else if you want to complete the square.

    thanks for put some info, do you know any good web sites for someone just starting off in algebra
    I edited my previous post after you edited yours to include the other info you needed to find.

    You could 'google' algebra tutorials or something like that to find some sites to help you. I found this one: algebasics™ Algebra Tutorials
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