## Calcmathlete 3 years ago 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).

1. TuringTest

my first thought is to try factoring and see if that takes us somewhere

2. Calcmathlete

Aren't they both unfactorable?

3. TuringTest

I guess, I'm just brainstorming (and multitasking)

4. TuringTest

changing the second into 2(x^2-4x+5) may shed some light is thought #2

5. TuringTest

let's see\[f(12)=102\le182\le202\]

6. TuringTest

hm...

7. TuringTest

I think I need to stop multitasking to do this

8. Calcmathlete

\[227 ≤ f(17) ≤ 462\]

9. TuringTest

I have a strong feeling that @asnaseer is going to save us

10. asnaseer

could we make use of the fact that if the LHS=RHS then that fixes f(x)?

11. asnaseer

i.e.\[x^2-4x+6=2x^2-8x+10\]f we solve this we get:\[(x-2)^2=0\implies x=2\]

12. asnaseer

so at x=2 both the LHS (left hand side) and RHS (right hand side) are equal

13. TuringTest

interesting...

14. Calcmathlete

That would make f(2) = 2?

15. Calcmathlete

When you plug it back in anyway...

16. asnaseer

yes

17. asnaseer

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

18. asnaseer

we have two equations so far

19. asnaseer

maybe make use of f(0)?

20. asnaseer

which gives us:\[6\le c\le 10\]

21. maheshmeghwal9

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

22. Calcmathlete

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?

23. asnaseer

f(1) is also interesting, it gives us:\[3\le a+b+c\le4\]

24. asnaseer

what do you mean by /strange/?

25. asnaseer

is there a particular approach that is used?

26. Calcmathlete

Well, the way they did it is a lot simpler and way different from this way...they completed the square...

27. asnaseer

hmmm... let me think over that approach...

28. maheshmeghwal9

Becoz we all are doing this type of question first time. & they had done it already. I really appreciate @asnaseer 's work:D

29. slaaibak

all three equations have a turning point (minimum) at x=2. so you could use that.

30. asnaseer

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:\[a+b+c=4\tag{3b}\]

31. Spacelimbus

Does it become a geometric question when they complete the square? |dw:1343584198825:dw|

32. Calcmathlete

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.

33. Spacelimbus

just thinking out loud, thought they might subtract squares then.

34. Spacelimbus

hmm no, that's not what I thought then.

35. Calcmathlete

I don't unserstand how they MUST have the same vertex though...

36. asnaseer

interesting observation @slaaibak - maybe we can use that to get the 3rd equation?

37. asnaseer

yes @Calcmathlete that ties up with @slaaibak's observation

38. slaaibak

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

39. maheshmeghwal9

ya so they all have same vertex. as @Calcmathlete 's.............. said:D

40. Calcmathlete

Thin is, I don't understand how they MUST all have the same vertex. Is it because of how the parabolas work when graphed?

41. asnaseer

the LHS and RHS have the same vertex which fixes the vertex of the middle one

42. maheshmeghwal9

no becoz they all have same minimum value at x=2

43. maheshmeghwal9

thus they all have same vertex

44. slaaibak

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

45. maheshmeghwal9

|dw:1343584535907:dw|

46. maheshmeghwal9

3 parabolas

47. maheshmeghwal9

48. asnaseer

very similar to the fact that x=2 gives the same value for LHS and RHS which fixes the value of f(2)

49. Calcmathlete

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?

50. slaaibak

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

51. asnaseer

yes @Calcmathlete

52. slaaibak

yeah that's correct

53. Calcmathlete

Alright. Thank you! I wish I could give multiple medals...

54. asnaseer

yw :) and yes, I wish we had Olympic-style medals - Gold/Silver/Bronze :)

55. asnaseer

but then we would have to drug test everyone to make sure they were not cheating! :D

56. Calcmathlete

lol...

57. maheshmeghwal9

Graphs must be there.