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Math Help - Quadratic function:convert to standard form?

  1. #1
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    Quadratic function:convert to standard form?

    Hi guys, i need help with these questions, i have attempted the questions myself and just wanted to double check my answers.
    convert to standard form
    1. y = 3x2 + 2x + 1 ( the answer i got is Y=-3(x+1/3)^2 -2/3) is this correct?
    2.
    y = 4(x + 5)(x 2) ( i can't figure how to solve this one)
    3. Y= -1/2 x^2 ( the answer i got is Y= 2(1-3)+2
    can someone please chck my answers and correct me i was wrong.
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  2. #2
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    Re: Quadratic function:convert to standard form?

    I suspect standard form means f(x)=a*x^2 + b*x +c
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    Re: Quadratic function:convert to standard form?

    Hello, Husky93!

    This is not Standard Form.

    You want the "Vertex Form": . y \:=\:a(x-h)^2 + k
    . . where the vertex is (h,k).


    Convert to Vertex Form.

    (1)\;y \:=\:3x^2 + 2x + 1

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

    y \;=\;-3\left(x^2 - \tfrac{2}{3}x \:{\color{red}+\: \tfrac{1}{9}} - \tfrac{1}{3} \:{\color{red}-\: \tfrac{1}{9}}\right)

    y \;=\;-3\left(\left[x - \tfrac{1}{3}\right]^2 - \tfrac{4}{9}\right)

    y \;=\;-3\left(x-\tfrac{1}{3}\right)^2 + \tfrac{4}{3}

    \text{Vertex: }\: \left(\tfrac{1}{3},\:\tfrac{4}{3}\right)




    (2)\;y \:=\: 4(x + 5)(x  2)

    y \;=\;4(x^2 + 3x - 10)

    y \;=\;4\left(x^2 + 3x\:{\color{red}+\:\tfrac{9}{4}} - 10\:{\color{red}-\:\tfrac{9}{4}}\right)

    y \;=\;4\left(\left[x+\tfrac{3}{2}\right]^2 - \tfrac{49}{4}\right)

    y \;=\; 4\left(x + \tfrac{3}{2}\right)^2 - 49

    \text{Vertex: }\:\left(\text{-}\tfrac{3}{2},\:\text{-}49\right)




    (3)\; y\:=\: -\tfrac{1}{2}x^2

    This is already in vertex form: . y \;=\;-\tfrac{1}{2}(x - 0)^2 + 0

    \text{Vertex: }\:(0,\,0)
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  4. #4
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    Re: Quadratic function:convert to standard form?

    Quote Originally Posted by Soroban View Post
    Hello, Husky93!

    This is not Standard Form.

    You want the "Vertex Form": . y \:=\:a(x-h)^2 + k
    . . where the vertex is (h,k).



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

    y \;=\;-3\left(x^2 - \tfrac{2}{3}x \:{\color{red}+\: \tfrac{1}{9}} - \tfrac{1}{3} \:{\color{red}-\: \tfrac{1}{9}}\right)

    y \;=\;-3\left(\left[x - \tfrac{1}{3}\right]^2 - \tfrac{4}{9}\right)

    y \;=\;-3\left(x-\tfrac{1}{3}\right)^2 + \tfrac{4}{3}

    \text{Vertex: }\: \left(\tfrac{1}{3},\:\tfrac{4}{3}\right)





    y \;=\;4(x^2 + 3x - 10)

    y \;=\;4\left(x^2 + 3x\:{\color{red}+\:\tfrac{9}{4}} - 10\:{\color{red}-\:\tfrac{9}{4}}\right)

    y \;=\;4\left(\left[x+\tfrac{3}{2}\right]^2 - \tfrac{49}{4}\right)

    y \;=\; 4\left(x + \tfrac{3}{2}\right)^2 - 49

    \text{Vertex: }\:\left(\text{-}\tfrac{3}{2},\:\text{-}49\right)





    This is already in vertex form: . y \;=\;-\tfrac{1}{2}(x - 0)^2 + 0

    0,\,0)" alt="\text{Vertex: }\0,\,0)" />
    The turning point form IS often referred as the Standard form, because it gives the most information.
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