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Math Help - Changing standard form to vertex form

  1. #1
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    Question Changing standard form to vertex form

    Please help me change this standard form of a parabola to vertex form.

    y = 0.2955x^2 + 0.1811x + 22.382

    If possible, please use the -b/2a version instead of completing the square, because that is what I was taught.

    Thank you for any help!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by live_laugh_luv27 View Post
    Please help me change this standard form of a parabola to vertex form.

    y = 0.2955x^2 + 0.1811x + 22.382

    If possible, please use the -b/2a version instead of completing the square, because that is what I was taught.

    Thank you for any help!
    completing the square for y = ax^2 + bx + c

    y = a \bigg( x^2 + \frac bax + \frac ca \bigg)

    \Rightarrow y = a \bigg[ x^2 + \frac bax + \left( \frac b{2a}\right)^2 - \left( \frac b{2a} \right)^2 + \frac ca \bigg]

    \Rightarrow y = a \bigg[ \left( x + \frac b{2a}\right)^2 + \frac {4ac - b^2}{4a^2} \bigg]

    \Rightarrow y = a \bigg(x + \frac b{2a} \bigg)^2 + \frac {4ac - b^2}{4a}
    Last edited by Jhevon; November 1st 2008 at 12:19 PM. Reason: forgot to square the 2, thanks Shyam
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  3. #3
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    Quote Originally Posted by live_laugh_luv27 View Post
    Please help me change this standard form of a parabola to vertex form.

    y = 0.2955x^2 + 0.1811x + 22.382

    If possible, please use the -b/2a version instead of completing the square, because that is what I was taught.

    Thank you for any help!
    y = a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}

    The vertex is \left(-\frac{b}{2a}, \;\frac{4ac-b^2}{4a}\right)

    here, a = 0.2955, b = 0.1811, c = 22.382

    y = 0.2955\left(x+\frac{0.1811}{2(0.2955)}\right)^2+\f  rac{4(0.2955)(22.382)-(0.1811)^2}{4(0.2955)}


    y = 0.2955\left(x+0.3064\right)^2+22.3542
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  4. #4
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    Quote Originally Posted by Shyam View Post
    y = a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}

    The vertex is \left(-\frac{b}{2a}, \;\frac{4ac-b^2}{4a}\right)

    here, a = 0.2955, b = 0.1811, c = 22.382

    y = 0.2955\left(x+\frac{0.1811}{2(0.2955)}\right)^2+\f  rac{4(0.2955)(22.382)-(0.1811)^2}{4(0.2955)}


    y = 0.2955\left(x+0.3064\right)^2+22.3542

    Thank you so much!!
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  5. #5
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    Question

    One question --
    One post has k= 2ac b2/2a, the other has k= 4ac b2/2a.
    Who is correct?
    Thanks!
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