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byndbtch

ax^2+bx+c<0 anyone have a clue as I dont!!

  • one year ago
  • one year ago

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  1. Mertsj
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    What are you supposed to do?

    • one year ago
  2. byndbtch
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    solve the equation or inequality???

    • one year ago
  3. byndbtch
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    there is a graph with it and I am so lost.....

    • one year ago
  4. byndbtch
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    I had ax^2+bx+c=0 and got that one, I think but this????

    • one year ago
  5. Mertsj
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    You need to post the entire problem including the graph. The only thing you posted in the general quadratic inequality. Its solution depends entirely upon the values of a, b, and c.

    • one year ago
  6. byndbtch
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    I know right....ok hang on. But I dont no how to post a graph??

    • one year ago
  7. Mertsj
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    Well then, you could tell what its shape is (I'm assuming it's a parabola), tell if it is concave upward or downward and give its vertex and x intercepts (if any). That would give us a pretty good idea of the problem.

    • one year ago
  8. byndbtch
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    it is upward, the vertex is 6.5 and -4 and 4 does that help

    • one year ago
  9. byndbtch
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    the vertex is -6

    • one year ago
  10. Mertsj
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    |dw:1349043948873:dw|

    • one year ago
  11. Mertsj
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    Does it look like that?

    • one year ago
  12. byndbtch
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    kinda but the vertex ir right in the middle.....it points upward at -4 on the left and 4 on the right

    • one year ago
  13. Mertsj
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    So the vertex is on the y axis?

    • one year ago
  14. byndbtch
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    yes

    • one year ago
  15. Mertsj
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    And it crosses the x axis at (4, 0) and (-4,0)?

    • one year ago
  16. byndbtch
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    yes

    • one year ago
  17. byndbtch
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    |dw:1349044244497:dw|

    • one year ago
  18. byndbtch
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    more like that

    • one year ago
  19. Mertsj
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    |dw:1349044229220:dw|

    • one year ago
  20. byndbtch
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    exactly!!

    • one year ago
  21. Mertsj
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    Now exactly what does the problem say to do?

    • one year ago
  22. byndbtch
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    The graph of y=ax^2+bx+c solve each equation . ax^2+bx+c=0 ax^2+bx+c<0 ax^2+bx+c>0

    • one year ago
  23. Mertsj
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    There must be more because what you posted doesn't make sense.

    • one year ago
  24. byndbtch
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    I know!! that is exactly as it is posted.

    • one year ago
  25. byndbtch
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    my options for the second equation are like 7 choices...makes no sense to me??

    • one year ago
  26. Mertsj
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    Does it say "The graph of y = x*2+bx+c is shown"?

    • one year ago
  27. byndbtch
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    YES

    • one year ago
  28. Mertsj
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    Ok. Now. I repeat my original request. Please post the problem EXACTLY as shown.

    • one year ago
  29. byndbtch
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    the graph of y=ax^2+bx+c is given below. Solve each equation or inequality. ax^2+bx+c=0 ax^2+bx+c<0 ax^2+bx+c>0

    • one year ago
  30. byndbtch
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    the vertex is not the -6 it is -30!!!!

    • one year ago
  31. byndbtch
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    its -32

    • one year ago
  32. byndbtch
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    I just seen that......does that help any???

    • one year ago
  33. Mertsj
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    What are the x intercepts?

    • one year ago
  34. byndbtch
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    they are the same

    • one year ago
  35. byndbtch
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    -4 4

    • one year ago
  36. Mertsj
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    Are you sure?

    • one year ago
  37. byndbtch
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    yes

    • one year ago
  38. byndbtch
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    |dw:1349045608333:dw|

    • one year ago
  39. Mertsj
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    Then plug into the equation y = a(x-h)^2 +k and find the value of a

    • one year ago
  40. Mertsj
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    Then solve the resulting inequality.

    • one year ago
  41. byndbtch
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    that is where I am confused....plug what numbers? and plug them in as a????

    • one year ago
  42. Mertsj
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    (h,k) is the vertex which in your case is (0,-32) You are given the x intercepts. Choose one of them, perhaps (4,0). Replace x with 4 and y with 0 and solve for a.

    • one year ago
  43. byndbtch
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    ok I will try. thx

    • one year ago
  44. Mertsj
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    yw

    • one year ago
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