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anonymous

  • 5 years ago

how to determine a & b valuesin the equation: y=ax^2 +bx-4 if the vertex is located at (-2,-1) Please help!

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  1. myininaya
    • 5 years ago
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    did you enter in -2 for x and -1 for y

  2. Hero
    • 5 years ago
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    4a = b

  3. Hero
    • 5 years ago
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    -1 = a(-2)^2 +(4a)(-2) - 4

  4. Hero
    • 5 years ago
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    a = -3/4?

  5. Hero
    • 5 years ago
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    b = 3?

  6. Hero
    • 5 years ago
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    How did you get b = 2?

  7. myininaya
    • 5 years ago
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    so a has different value too

  8. Hero
    • 5 years ago
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    myininaya, how come you just can't do things the normal way?

  9. anonymous
    • 5 years ago
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    so you cant just sub -2 into one equation and -1 into another, andthe other number equal zero? or does that not mke any sense?

  10. anonymous
    • 5 years ago
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    I'm just a little confused by the math you did above..lol

  11. Hero
    • 5 years ago
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    myininaya likes to go overboard with her methods

  12. myininaya
    • 5 years ago
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    ok hero but there are two possible outcomes

  13. Hero
    • 5 years ago
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    Not if you do it the way I did. I used direct substitution

  14. myininaya
    • 5 years ago
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    We have a parabola concave up and also concave down at vertex

  15. myininaya
    • 5 years ago
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    Here I will do it the way I told you to

  16. anonymous
    • 5 years ago
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    Im familar with substitution , just not the way shown above im only in gr 10 so mabye thats why i dont get yours?

  17. Hero
    • 5 years ago
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    a = -3/4 b = -3

  18. anonymous
    • 5 years ago
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    ^did you make 2 equations and then substitutions?

  19. Hero
    • 5 years ago
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    something like that

  20. anonymous
    • 5 years ago
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    -1 = a(-2)^2 +(4a)(-2) - 4 :this is what you did, why did you sub 4a into the equation? is there a more strait forward way to do this? :)

  21. Hero
    • 5 years ago
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    myininaya, if you use the direct substitution method, you wouldn't have to worry about getting hung up like that. The parabola opens upward or downward depending on the value of a. Since I calculated a to be negative, I'm pretty sure it opens downward and there is only one possibility

  22. Hero
    • 5 years ago
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    I replace b with 4a because b = 4a

  23. Hero
    • 5 years ago
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    I also replaced x = -2 and y = -1

  24. Hero
    • 5 years ago
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    Which are all of the known values

  25. Hero
    • 5 years ago
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    according to x = -b/2a, there's only one possible a

  26. anonymous
    • 5 years ago
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    alright im still a bit confused but thanks for your help!! :)

  27. myininaya
    • 5 years ago
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    \[-1=a(-2)^2+b(-2)-4\] \[-1=4a-2b-4 => 3=4a-2b\] We also know the vertex which is (-2,-1) \[y=a(x-h)^2+k\] we also know h=-2 and k=-1 so we have \[y=a(x+2)^2-1=a(x^2+4x+4)-1=ax^2+4ax+4a-1=ax^2+4ax+(4a-1)\] \[=>4a=b , -4=4a-1\] \[=>3=b-2b=> 3=-b=> b=-3\] if b=-3 then a=-3/4 hmmm how do we get the other possibility ok so maybe there is only one lol

  28. Hero
    • 5 years ago
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    :D

  29. Hero
    • 5 years ago
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    There's only one possibility because there's only one a value that would work in order to get both sides equal for x = -b/2a

  30. Hero
    • 5 years ago
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    if a was squared, then there would be two possibilities

  31. myininaya
    • 5 years ago
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    I didn't think about the y-intercept

  32. myininaya
    • 5 years ago
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    The parabola has vertex (-2,-1) and -4 is lower so this graph will be concave down

  33. myininaya
    • 5 years ago
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    |dw:1327278866372:dw|

  34. Hero
    • 5 years ago
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    yup

  35. myininaya
    • 5 years ago
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    we can't not have it concave up is has to come to that -4

  36. myininaya
    • 5 years ago
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    so my bad hero you win this one

  37. Hero
    • 5 years ago
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    I don't win many so I will savor this :P

  38. myininaya
    • 5 years ago
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    hey lady do you understand what i did up there?

  39. anonymous
    • 5 years ago
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    A bit lol but thanks!

  40. Hero
    • 5 years ago
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    mya likes to confuse people then ask if they understand

  41. myininaya
    • 5 years ago
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    you also have to use the vertex form like i did to find another equation or two

  42. anonymous
    • 5 years ago
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    im still pretty confused but i can move on to a different qestion on my assignment lol thanks for the help on this one tho :)

  43. Hero
    • 5 years ago
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    You're still here?

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