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anonymous

  • one year ago

Find a quadratic model for the set of values: (-2, -20), (0, -4), (4,-20) Show your work

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  1. anonymous
    • one year ago
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    @UsukiDoll

  2. perl
    • one year ago
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    You need a system of equations

  3. UsukiDoll
    • one year ago
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    I don't feel well today. I woke up with a migraine (again)

  4. anonymous
    • one year ago
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    Take some advil and drink water

  5. perl
    • one year ago
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    $$ \Large{ y = ax^2 + bx + c \\~\\ -20 = a(-2)^2 + b(-2) + c \\-4 = a(0)^2 + b(0) + c \\ -20 = a(4)^2 + b(4) + c }$$

  6. anonymous
    • one year ago
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    i honestly have no clue how to do this equation

  7. perl
    • one year ago
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    Do you agree with what I have so far. I plugged in the values into a generic quadratic function.

  8. anonymous
    • one year ago
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    I think so, i wouldn't know if you were wrong

  9. perl
    • one year ago
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    Finding a quadratic model means fitting the points to a quadratic curve \( \large y = ax^2 + bx + c \) . So our job is to find the parameters \( \large a,b,c\)

  10. anonymous
    • one year ago
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    @perl how do i find the a,b,c

  11. perl
    • one year ago
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    We have these three equations so far. \[ \\~\\\Large { -20 = a(-2)^2 + b(-2) + c \\-4 = a(0)^2 + b(0) + c \\ -20 = a(4)^2 + b(4) + c }\]

  12. perl
    • one year ago
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    the second equation gives us -4 = 0 + 0 + c -4 = c

  13. anonymous
    • one year ago
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    So c=-4

  14. perl
    • one year ago
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    Yes. Now back substitute.

  15. perl
    • one year ago
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    We have these three equations so far. \[ \\~\\\Large { -20 = a(-2)^2 + b(-2) + (-4) \\-4 = a(0)^2 + b(0) + (-4) \\ -20 = a(4)^2 + b(4) + (-4) }\]

  16. perl
    • one year ago
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    we don't need the second equation any longer

  17. perl
    • one year ago
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    \[ \\~\\\Large { -20 = a(-2)^2 + b(-2) + (-4) \\ -20 = a(4)^2 + b(4) + (-4) }\]

  18. anonymous
    • one year ago
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    im writing as we go

  19. perl
    • one year ago
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    \[ \\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = 16a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = 16a -2b \\ -16 = 16a + 4b }\]

  20. anonymous
    • one year ago
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    Is that the final answer?

  21. perl
    • one year ago
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    We can solve for `b`

  22. anonymous
    • one year ago
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    So we haven't solved A or C yet?

  23. perl
    • one year ago
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    we solved for c = -4

  24. anonymous
    • one year ago
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    Ok so how do we solve for A and B than? Im sorry i really dont understand

  25. anonymous
    • one year ago
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    @perl

  26. perl
    • one year ago
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    we can solve that linear equation

  27. anonymous
    • one year ago
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    how?

  28. perl
    • one year ago
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    \[ \\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = 16a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = 16a -2b \\ -16 = 16a + 4b \\~\\ \rm add ~ the ~ two ~ equations\\~\\ \\-32 = 2b \\~\\ \\b = \frac{-32}{2} }\]

  29. anonymous
    • one year ago
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    So b=-16?

  30. perl
    • one year ago
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    yes

  31. anonymous
    • one year ago
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    So now we need to find A

  32. perl
    • one year ago
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    y = -2x^2 + 4x -4 should be the answer. I have an error somewhere above.

  33. anonymous
    • one year ago
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    so b should really be b=4?

  34. anonymous
    • one year ago
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    A=-2 B=4 C=-4?

  35. perl
    • one year ago
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    yes

  36. perl
    • one year ago
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    I have to go back and find the error now ;)

  37. anonymous
    • one year ago
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    Awesome thank you so much!

  38. perl
    • one year ago
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    but this method is sound if you want to try it again

  39. perl
    • one year ago
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    if you ever*

  40. anonymous
    • one year ago
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    yeah i prob will, i have a whole other algebra class ahead of me

  41. perl
    • one year ago
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    \[ \\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = \color{red}4a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = \color{red}4a -2b \\ -16 = 16a + 4b \\~\\ \rm solve ~this~ system~ of~ equations\\~\\ \\ a = -2 ~ , b = 4 \\~\\ \rm backsubstitute \\y = -2x^2 + 4x - 4 }\]

  42. perl
    • one year ago
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    Another way to solve this problem is to note that your parabola has a y intercept of -4, because x=0 at the y intercept. That reduces the problem of solving a linear system of 2 equations as we have above.

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