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anonymous

  • 5 years ago

How do I get the quadratic root equation numbers from a graph?

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  1. anonymous
    • 5 years ago
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    well what does it mean to have the root of an equation? it means you have the 0s right?

  2. anonymous
    • 5 years ago
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    i know how to balance, but how do I get the ax2+bx+c #s from it?

  3. anonymous
    • 5 years ago
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    well first thing you want to do is find where it intersects the axis correct? where it intersects is going to give you your roots. so you will have something like (x+5)(x-3)

  4. anonymous
    • 5 years ago
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    then you can foil that guy out

  5. anonymous
    • 5 years ago
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    oh, the points on the x-axis incept are the roots?

  6. anonymous
    • 5 years ago
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    well isnt that how we defined the quadratic equation? the quadratic equation is set up such that your solving for the left hand side to equal 0

  7. anonymous
    • 5 years ago
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    so if i have a graph and the two parabola intercepts are 2 and -2, then are those the roots, or are those the x +2 and x-2, or do they fit into the ax2+bx+c?

  8. anonymous
    • 5 years ago
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    on the x-axis

  9. anonymous
    • 5 years ago
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    oh you have multiple graphs?

  10. anonymous
    • 5 years ago
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    no i have one graph, it's a parabola that intercepts the x-axis twice

  11. anonymous
    • 5 years ago
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    whats the equation of a parabola? the general syntax? that should help you quite a bit if you analyze what happens with the parabola equation

  12. anonymous
    • 5 years ago
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    i just don't know the equation for turning graph data into the qudratic root formula

  13. anonymous
    • 5 years ago
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    ok well think about (x+2)(x-2)

  14. anonymous
    • 5 years ago
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    that goes to x^2 - 4 correct?

  15. anonymous
    • 5 years ago
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    your graph intercepts the y axis at -4 im guessing?

  16. anonymous
    • 5 years ago
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    well.. you can change where it intercepts the y axis

  17. anonymous
    • 5 years ago
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    is that "c", where it intercepts?

  18. anonymous
    • 5 years ago
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    idk the equation for the parabola im justi thinking of how the graph works out intuitevely

  19. anonymous
    • 5 years ago
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    i mean i have the equation. but think about it like this. (x^2 - 4)/c

  20. anonymous
    • 5 years ago
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    the c is going to effect the 'width' of the graph

  21. anonymous
    • 5 years ago
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    the x and y intercepts are still at 2 and -2

  22. anonymous
    • 5 years ago
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    those are both the x intercepts the y-one is -4

  23. anonymous
    • 5 years ago
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    but the c is going to change where it intercepts the y axis. you can solve for what c has to be though for the y intercept.

  24. anonymous
    • 5 years ago
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    y = (x^2-4)/C

  25. anonymous
    • 5 years ago
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    Set x to 0 and y to -4

  26. anonymous
    • 5 years ago
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    solve for C

  27. anonymous
    • 5 years ago
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    I know c=y-intercept, right? do i still use ax2... etc etc or do i use y=ax2 + c and how do i get "a"

  28. anonymous
    • 5 years ago
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    well in my example my 'a' would be 1/C

  29. anonymous
    • 5 years ago
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    and your 'c' would be my 4/C

  30. anonymous
    • 5 years ago
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    so y = 1X^2-4 where a = 1 and C = -4...

  31. anonymous
    • 5 years ago
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    X intercepts are the roots?

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