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Calcmathlete Group Title

Let f(x) be a quadratic function. If \(x^2 - 4x + 6 ≤ f(x) ≤ 2x^2 - 8x +10\) for all real numbers x, and f(12) = 182, find f(17).

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    my first thought is to try factoring and see if that takes us somewhere

    • 2 years ago
  2. Calcmathlete Group Title
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    Aren't they both unfactorable?

    • 2 years ago
  3. TuringTest Group Title
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    I guess, I'm just brainstorming (and multitasking)

    • 2 years ago
  4. TuringTest Group Title
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    changing the second into 2(x^2-4x+5) may shed some light is thought #2

    • 2 years ago
  5. TuringTest Group Title
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    let's see\[f(12)=102\le182\le202\]

    • 2 years ago
  6. TuringTest Group Title
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    hm...

    • 2 years ago
  7. TuringTest Group Title
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    I think I need to stop multitasking to do this

    • 2 years ago
  8. Calcmathlete Group Title
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    \[227 ≤ f(17) ≤ 462\]

    • 2 years ago
  9. TuringTest Group Title
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    I have a strong feeling that @asnaseer is going to save us

    • 2 years ago
  10. asnaseer Group Title
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    could we make use of the fact that if the LHS=RHS then that fixes f(x)?

    • 2 years ago
  11. asnaseer Group Title
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    i.e.\[x^2-4x+6=2x^2-8x+10\]f we solve this we get:\[(x-2)^2=0\implies x=2\]

    • 2 years ago
  12. asnaseer Group Title
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    so at x=2 both the LHS (left hand side) and RHS (right hand side) are equal

    • 2 years ago
  13. TuringTest Group Title
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    interesting...

    • 2 years ago
  14. Calcmathlete Group Title
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    That would make f(2) = 2?

    • 2 years ago
  15. Calcmathlete Group Title
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    When you plug it back in anyway...

    • 2 years ago
  16. asnaseer Group Title
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    yes

    • 2 years ago
  17. asnaseer Group Title
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    and if we define f(x) as:\[f(x)=ax^2+bx+c\]then maybe we can form 3 equations and solve for a, b and c

    • 2 years ago
  18. asnaseer Group Title
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    we have two equations so far

    • 2 years ago
  19. asnaseer Group Title
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    maybe make use of f(0)?

    • 2 years ago
  20. asnaseer Group Title
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    which gives us:\[6\le c\le 10\]

    • 2 years ago
  21. maheshmeghwal9 Group Title
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    \[x^2 - 4x + 6 ≤ (ax^2+bx+c)≤ 2x^2 - 8x +10\] \[\color{blue}{a(12)^2+b(12)+c}=182=\color{blue}{144a+12b+c.}\] \[\color{red}{a(17)^2+b(17)+c=?}\]\[6 \le c \le 10.\]

    • 2 years ago
  22. Calcmathlete Group Title
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    I just looked up how they did it...it's very strange, and I'm having a bit of trouble understanding a part of it...do you want me to post it or would you like to try figuring the rest out?

    • 2 years ago
  23. asnaseer Group Title
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    f(1) is also interesting, it gives us:\[3\le a+b+c\le4\]

    • 2 years ago
  24. asnaseer Group Title
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    what do you mean by /strange/?

    • 2 years ago
  25. asnaseer Group Title
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    is there a particular approach that is used?

    • 2 years ago
  26. Calcmathlete Group Title
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    Well, the way they did it is a lot simpler and way different from this way...they completed the square...

    • 2 years ago
  27. asnaseer Group Title
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    hmmm... let me think over that approach...

    • 2 years ago
  28. maheshmeghwal9 Group Title
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    Becoz we all are doing this type of question first time. & they had done it already. I really appreciate @asnaseer 's work:D

    • 2 years ago
  29. slaaibak Group Title
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    all three equations have a turning point (minimum) at x=2. so you could use that.

    • 2 years ago
  30. asnaseer Group Title
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    I can't /see/ how completing the square helps. but I wonder if we can make use of my last result for f(1) to get 3 possible equations to solve.\[144a+12b+c=182\tag{1}\]\[4a+2b+c=2\tag{2}\]and then either:\[a+b+c=3\tag{3a}\]or:\[a+b+c=4\tag{3b}\]

    • 2 years ago
  31. Spacelimbus Group Title
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    Does it become a geometric question when they complete the square? |dw:1343584198825:dw|

    • 2 years ago
  32. Calcmathlete Group Title
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    They said that once you complete the square, you find that f(x) must have the same vertex as the other two equations and that you just have to solve for a.

    • 2 years ago
  33. Spacelimbus Group Title
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    just thinking out loud, thought they might subtract squares then.

    • 2 years ago
  34. Spacelimbus Group Title
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    hmm no, that's not what I thought then.

    • 2 years ago
  35. Calcmathlete Group Title
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    I don't unserstand how they MUST have the same vertex though...

    • 2 years ago
  36. asnaseer Group Title
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    interesting observation @slaaibak - maybe we can use that to get the 3rd equation?

    • 2 years ago
  37. asnaseer Group Title
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    yes @Calcmathlete that ties up with @slaaibak's observation

    • 2 years ago
  38. slaaibak Group Title
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    if we use f(x) = a(x-b)^2 + k, we can solve. because we have b:2 and k can be found by f(2) = k so it's pretty easy from here

    • 2 years ago
  39. maheshmeghwal9 Group Title
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    ya so they all have same vertex. as @Calcmathlete 's.............. said:D

    • 2 years ago
  40. Calcmathlete Group Title
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    Thin is, I don't understand how they MUST all have the same vertex. Is it because of how the parabolas work when graphed?

    • 2 years ago
  41. asnaseer Group Title
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    the LHS and RHS have the same vertex which fixes the vertex of the middle one

    • 2 years ago
  42. maheshmeghwal9 Group Title
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    no becoz they all have same minimum value at x=2

    • 2 years ago
  43. maheshmeghwal9 Group Title
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    thus they all have same vertex

    • 2 years ago
  44. slaaibak Group Title
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    both have the same turning point. therefore, f(x) can't be decreasing if the other two functions are increasing, because f(2) are the same for all the equations. therefore all of them have a minimum at 2

    • 2 years ago
  45. maheshmeghwal9 Group Title
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    |dw:1343584535907:dw|

    • 2 years ago
  46. maheshmeghwal9 Group Title
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    3 parabolas

    • 2 years ago
  47. maheshmeghwal9 Group Title
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    3 functions quadratic......

    • 2 years ago
  48. asnaseer Group Title
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    very similar to the fact that x=2 gives the same value for LHS and RHS which fixes the value of f(2)

    • 2 years ago
  49. Calcmathlete Group Title
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    Alright. That's really why i didn't understand. I guess since the opposite ends of the compound inequality have the same vertex, it is justified. From there it is easy I guess. \[f(12) = a(x - 2)^2 + 2 \implies a = 1.8\]Then we can just plug in 17? \[f(17) = 1.8(225) + 2 = 405 Is that right?

    • 2 years ago
  50. slaaibak Group Title
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    f(2) = 2 f(x) = a(x-2)^2 + k b=2 due to minimum so f(2) = a(2-2)^2 + k = 2 therefore k = 2 now f(12) = 182 so 182 = a(12-2)^2 + 2 182 = 100a + 2 100a = 180 a = 1.8

    • 2 years ago
  51. asnaseer Group Title
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    yes @Calcmathlete

    • 2 years ago
  52. slaaibak Group Title
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    yeah that's correct

    • 2 years ago
  53. Calcmathlete Group Title
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    Alright. Thank you! I wish I could give multiple medals...

    • 2 years ago
  54. asnaseer Group Title
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    yw :) and yes, I wish we had Olympic-style medals - Gold/Silver/Bronze :)

    • 2 years ago
  55. asnaseer Group Title
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    but then we would have to drug test everyone to make sure they were not cheating! :D

    • 2 years ago
  56. Calcmathlete Group Title
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    lol...

    • 2 years ago
  57. maheshmeghwal9 Group Title
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    Graphs must be there.

    • 2 years ago
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