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Math Help - quadratic function standard form

  1. #1
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    quadratic function standard form

    Ok here some excercise from the book that I don't get it because the book the does not have an example how to do this.

    f(x)= .5x^2-.5

    I couldn't figure out how to transform this to standard form of quadratic function from general form.

    please anyone can explain how.... thanks!
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  2. #2
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    Quote Originally Posted by Anemori View Post
    Ok here some excercise from the book that I don't get it because the book the does not have an example how to do this.

    f(x)= .5x^2-.5

    I couldn't figure out how to transform this to standard form of quadratic function from general form.

    please anyone can explain how.... thanks!
    Take out the common factor of 0.5.

    f(x) = 0.5(x^2 - 1)

     = 0.5[(x - 0)^2 - 1]

     = 0.5(x - 0)^2 - 0.5
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Take out the common factor of 0.5.

    f(x) = 0.5(x^2 - 1)

     = 0.5[(x - 0)^2 - 1]

     = 0.5(x - 0)^2 - 0.5

    Thanks. I'm right with my guess then hehehe...

    what I did is that I did not transform it to standard form.

    f(x) = .5x^2-.5

    a= .5 , b= 0 , c= .5

    x= -b/2(a) ==> 0/2(.5) =0

    x= 0

    f(0) = .5(0)^2-.5 = -.5

    y = -.5

    vertex point of the graph is (0, -.5)
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  4. #4
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    That should have been obvious immediately.

    Since a square is never negative, .5x^2 is never negative so y= .5+ .5x^2 is ".5 plus something". That is, y is never less than .5 and will be equal to .5 only when x= 0. The vertex, the lowest point on the graph, is at (0, .5).

    (Don't just memorize formulas. Think about what you are doing!)
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