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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. sdfgsdfgs
    • one year ago
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    Since u need to show ur work....the general quadratic model is y=ax^2 + bx + c Now we need to find the constants a, b, c ok?

  2. anonymous
    • one year ago
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    How do we do that

  3. sdfgsdfgs
    • one year ago
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    U are given 3 points: (-2, -20), (0, -4), (4, -20) So u can use them to solve for abc

  4. anonymous
    • one year ago
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    I suck at this and I don't get it

  5. sdfgsdfgs
    • one year ago
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    I will do the first one for u, @Raylynngrace :) y=ax^2 + bx + c for (-2, -20) so -20 = a*(-2)^2 + b*(-2) + c -20 = 4a - 2b + c ok?

  6. anonymous
    • one year ago
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    Okay I get that now lol

  7. sdfgsdfgs
    • one year ago
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    Cool! Show me how u set up with the other two given points plz?

  8. anonymous
    • one year ago
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    idk if I did this right lol but for the next one I got y=ax^2+bx+c -4=a*(0)^2+b*(0) -4=a-b+c

  9. sdfgsdfgs
    • one year ago
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    good try but u missed something...it should be like -4=a*(0)^2+b*(0) + c -4=0+0+c c=-4 hey we found c already!

  10. anonymous
    • one year ago
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    Least I tried lol can you help me with the last one

  11. sdfgsdfgs
    • one year ago
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    nope u do it plz? i will help of cus :)

  12. anonymous
    • one year ago
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    y=ax^2+bx+c -20=a*(4)^2+b*(4) -20=16a+4b

  13. anonymous
    • one year ago
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    Did I get this one wrong to @sdfgsdfgs

  14. sdfgsdfgs
    • one year ago
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    @raylynngrace hahahaa u keep forgetting the c term....so it should be: -20=a*(4)^2+b*(4)+c -20=16a+4b+c now u got c=-4 already and from the first point: -20 = 4a - 2b + c U can solve for a and b :)

  15. anonymous
    • one year ago
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    @Raylynngrace This will help you a bit. Try !! http://www.softschools.com/math/algebra/quadratic_functions/quadratic_function_with_three_points/

  16. anonymous
    • one year ago
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    Thank you both for your help @sdfgsdfgs @JoshKoikkara

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