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briana.img

  • one year ago

Can someone help me learn Graphing a Parabola with a Vertex at the Origin?

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  1. briana.img
    • one year ago
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    My online class isn't explaining it well and I can't find videos with the problems I'm faced with.

  2. briana.img
    • one year ago
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    These are one of the examples

  3. Mr_Perfection_xD
    • one year ago
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    Phantom Lord got this.

  4. briana.img
    • one year ago
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  5. nincompoop
    • one year ago
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    first, covert it into either \(Ax^2 + bx + c = 0 \) or \(f(x) = y = a(x-h)^2 +k \) where \(h, k \) is your vertex and \(a\neq 0\) the first option is easier, the second option requires you to know how to complete the square

  6. nincompoop
    • one year ago
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    you can solve the vertex in the first one by using \(\large (x=\frac{-b}{2a},y = f(x)) \)

  7. briana.img
    • one year ago
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    @nincompoop how would you put y=1/12x^2 in that first formula???

  8. nincompoop
    • one year ago
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    so if you have a standard form of \(y = 2x^2 - 4x + 3\) your vertex can be solved by identifying your coefficients a = 2 b = 4 c = 3 vertex \(\large \frac{-b}{2a} = \frac{-(2)}{2(4)} = -\frac{1}{4}\) then plug \(- \frac{1}{4} \) into all of the x in \(y = 2x^2+4x+3 \)

  9. briana.img
    • one year ago
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    @nincompoop i think i'll just keep trying with youtube because you're confusing me even more

  10. nincompoop
    • one year ago
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    analyze the standard form \(y = Ax^2 + Bx +c \) you have three terms first term \(Ax^2 \) second term \(Bx \) third term \(c \)

  11. nincompoop
    • one year ago
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    ask yourself if the given equation has second or third term? if they don't show up it means it is zero anything +0 will remain the same

  12. nincompoop
    • one year ago
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    http://www.mathsisfun.com/geometry/parabola.html

  13. nincompoop
    • one year ago
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    http://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html

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