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anonymous

  • 5 years ago

the function f is defined for all real numbers x by f(x) = ax^2 + bx + c where a, b and c are constants and a is negative, in the xy-plane, the x coordinate of the vertex of the parabola is y=f(x) is -1. if t is a number for which f(t) > f(0), which of the following (any or all) must be true? 1: -2 < t < 0 2: f(t) < f(-2) 3: f(t) > f(1) please help!

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  1. myininaya
    • 5 years ago
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    we know this parabola is oven downward because a is negative and the vertex is (-1,k)

  2. myininaya
    • 5 years ago
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    f(t)>f(0) ....

  3. myininaya
    • 5 years ago
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    f(t)=a(t+1)^2+k and f(0)=a(0+1)^2+k=a+k and f(t)>f(0) so we have a(t+1)^2+k=a+k

  4. myininaya
    • 5 years ago
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    solve for t

  5. myininaya
    • 5 years ago
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    so -2<t<0

  6. anonymous
    • 5 years ago
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    so are you solving as a system here?

  7. myininaya
    • 5 years ago
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    i solve that equation by using remedial algebra

  8. myininaya
    • 5 years ago
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    subtract k on both sides

  9. myininaya
    • 5 years ago
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    divide by a on both sides

  10. anonymous
    • 5 years ago
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    so (t+1)^2 = 1

  11. myininaya
    • 5 years ago
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    yes

  12. myininaya
    • 5 years ago
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    square root both sides

  13. myininaya
    • 5 years ago
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    you get t+1=+ or - 1

  14. myininaya
    • 5 years ago
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    t=-1-1,1-1=-2,0

  15. anonymous
    • 5 years ago
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    ok, i think i understand that. also, it says option 3 is true, can you help with that?

  16. myininaya
    • 5 years ago
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    lets see we gots to think I can't believe I could actually about the first one lol

  17. myininaya
    • 5 years ago
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    i have a question where i have equal sign, it should had been a greater than sign a(t+1)^2+k>a+k I don't know if you caught that or not

  18. anonymous
    • 5 years ago
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    yes, i assumed it was since the problem is dealing with inequality

  19. myininaya
    • 5 years ago
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    okay cool im still thinking about 3

  20. myininaya
    • 5 years ago
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    i wonder what happens if we assume f(t)>f(1) and solve for t again im going to try

  21. myininaya
    • 5 years ago
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    we get -3<t<1

  22. myininaya
    • 5 years ago
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    you said the last one is true?

  23. myininaya
    • 5 years ago
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    the interval (-3,1) is not included in the first interval we got so i would think the 3rd one would be false unless i made a mistake somewhere

  24. anonymous
    • 5 years ago
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    right, but possibly the range works somehow? i have the harder time getting why option 3 works...

  25. myininaya
    • 5 years ago
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    oh wait we have t>1 or t<-3 not -3<t<1

  26. myininaya
    • 5 years ago
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    oomg wait are you for sure number 1 is right? is it true or false?

  27. myininaya
    • 5 years ago
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    ok sqrt(x^2)=|x|, so we have sqrt((t+1)^2)>1 and sqrt((t+1)^2))<-1 so we have t>0 or t<-2 and for the last one we got t<-3 or t>1 which are included in that interval so number 3 is true and number 1 is false now what about number 2...

  28. myininaya
    • 5 years ago
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    for the second one we get -2<t<0 but we said t>0 or t<-2 so number 2 is false

  29. myininaya
    • 5 years ago
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    i get it. the only true one is number 3 the others are FALSE!

  30. myininaya
    • 5 years ago
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    what do you think? any questions/

  31. myininaya
    • 5 years ago
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    i'm fixing to go to bed if you dont have any questions

  32. myininaya
    • 5 years ago
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    I really don't want to leave if you have questions, but I'm sleepy and I don't know if I should wait or not

  33. anonymous
    • 5 years ago
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    i think i kind of get it, although i'm not sure where the absolute value comes in

  34. myininaya
    • 5 years ago
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    sqrt(x^2)=|x| just think about it whats sqrt((-3)^2) and is equal to |-3|

  35. myininaya
    • 5 years ago
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    since I had something square and i took the square root of it that made me think of the absolute value thing where you change the sign of the number when the inequality sign is flipped

  36. anonymous
    • 5 years ago
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    ohh, ok that's right, i think i understand now, that was the one part that i didn't follow

  37. anonymous
    • 5 years ago
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    thank you so much for your help!

  38. myininaya
    • 5 years ago
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    the first time i did part one i got the interval wrong i don't know if you notice i corrected myself on that one or not

  39. anonymous
    • 5 years ago
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    yes, i saw, thank you for checking

  40. myininaya
    • 5 years ago
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    we know that interval is true since the question said if it was true right?

  41. myininaya
    • 5 years ago
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    so we based everything off that result

  42. anonymous
    • 5 years ago
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    exactly, i think we have to assume that base on the problem

  43. myininaya
    • 5 years ago
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    yes we do because the question says too

  44. myininaya
    • 5 years ago
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    i really like that problem that was totally awesome goodnight!

  45. anonymous
    • 5 years ago
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    thanks so much for your help, good night to you too!

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