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byndbtch

  • 2 years ago

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

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  1. Mertsj
    • 2 years ago
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    What are you supposed to do?

  2. byndbtch
    • 2 years ago
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    solve the equation or inequality???

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

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

  5. Mertsj
    • 2 years ago
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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.

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

  7. Mertsj
    • 2 years ago
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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.

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

  9. byndbtch
    • 2 years ago
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    the vertex is -6

  10. Mertsj
    • 2 years ago
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    |dw:1349043948873:dw|

  11. Mertsj
    • 2 years ago
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    Does it look like that?

  12. byndbtch
    • 2 years ago
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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

  13. Mertsj
    • 2 years ago
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    So the vertex is on the y axis?

  14. byndbtch
    • 2 years ago
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    yes

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

  16. byndbtch
    • 2 years ago
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    yes

  17. byndbtch
    • 2 years ago
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    |dw:1349044244497:dw|

  18. byndbtch
    • 2 years ago
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    more like that

  19. Mertsj
    • 2 years ago
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    |dw:1349044229220:dw|

  20. byndbtch
    • 2 years ago
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    exactly!!

  21. Mertsj
    • 2 years ago
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    Now exactly what does the problem say to do?

  22. byndbtch
    • 2 years ago
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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

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

  24. byndbtch
    • 2 years ago
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    I know!! that is exactly as it is posted.

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

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

  27. byndbtch
    • 2 years ago
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    YES

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

  29. byndbtch
    • 2 years ago
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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

  30. byndbtch
    • 2 years ago
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    the vertex is not the -6 it is -30!!!!

  31. byndbtch
    • 2 years ago
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    its -32

  32. byndbtch
    • 2 years ago
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    I just seen that......does that help any???

  33. Mertsj
    • 2 years ago
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    What are the x intercepts?

  34. byndbtch
    • 2 years ago
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    they are the same

  35. byndbtch
    • 2 years ago
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    -4 4

  36. Mertsj
    • 2 years ago
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    Are you sure?

  37. byndbtch
    • 2 years ago
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    yes

  38. byndbtch
    • 2 years ago
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    |dw:1349045608333:dw|

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

  40. Mertsj
    • 2 years ago
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    Then solve the resulting inequality.

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

  42. Mertsj
    • 2 years ago
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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.

  43. byndbtch
    • 2 years ago
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    ok I will try. thx

  44. Mertsj
    • 2 years ago
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    yw

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