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keenaH

  • 4 years ago

y=1/3x^2-4x+6

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  1. wombat
    • 4 years ago
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    Solve for x?

  2. keenaH
    • 4 years ago
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    im suposed to graph it

  3. wombat
    • 4 years ago
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    Ok. Well automatically, you can tell that it is a parabola (quadratic) right? Quadratics are more easily graphed when you have the equation in quadratic form: y=a(x-h)^2 +k So you will want to convert the above equation to this form. They you can take the vertex and y intercept from the form

  4. keenaH
    • 4 years ago
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    i forgot the step after finding the vertex and the formula you use to do it

  5. wombat
    • 4 years ago
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    The vertex is the point (h,k)

  6. keenaH
    • 4 years ago
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    im pretty sure the vertex is(6,-6)

  7. keenaH
    • 4 years ago
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    i used the formula ...x=-(b)/2(a)

  8. wombat
    • 4 years ago
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    That formula works too :) That will give you the x value that is the axis of symmetry. In the case you are doing, the vertex will be ( [-b/2a], f([-b/2a]))

  9. keenaH
    • 4 years ago
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    its so confusing

  10. keenaH
    • 4 years ago
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    i dont know what you just wrote, its like a foriegn language

  11. keenaH
    • 4 years ago
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    ...

  12. wombat
    • 4 years ago
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    OH! sorry! um what I wrote represents the coordinate pair of the vertex. -b/2a is the x value. f(-b/2a) is the y value that coresponds with the above x value

  13. afranklin12
    • 4 years ago
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    i got 2.67

  14. jhonyy9
    • 4 years ago
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    - for you can solve it you need make it equal zero - so 1/3 x^2 -4x +6 =0 multiply both sides by 3 3x^2 ---- -12x +18 =0 3 x^2 -12x +18 =0 D =144 - 4*1*18 =144 -72 =72 x_1,2= (12+/- sqrt72)/2 =(12+/- sqrt(9*4*2))/2 =(12+/- 6sqrt2)/2 = =2(6+/-3sqrt2)/2 =6+/-3sqrt2 x_1,2=6+/-3sqrt2

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