anonymous
  • anonymous
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).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
my first thought is to try factoring and see if that takes us somewhere
anonymous
  • anonymous
Aren't they both unfactorable?
TuringTest
  • TuringTest
I guess, I'm just brainstorming (and multitasking)

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TuringTest
  • TuringTest
changing the second into 2(x^2-4x+5) may shed some light is thought #2
TuringTest
  • TuringTest
let's see\[f(12)=102\le182\le202\]
TuringTest
  • TuringTest
hm...
TuringTest
  • TuringTest
I think I need to stop multitasking to do this
anonymous
  • anonymous
\[227 ≤ f(17) ≤ 462\]
TuringTest
  • TuringTest
I have a strong feeling that @asnaseer is going to save us
asnaseer
  • asnaseer
could we make use of the fact that if the LHS=RHS then that fixes f(x)?
asnaseer
  • asnaseer
i.e.\[x^2-4x+6=2x^2-8x+10\]f we solve this we get:\[(x-2)^2=0\implies x=2\]
asnaseer
  • asnaseer
so at x=2 both the LHS (left hand side) and RHS (right hand side) are equal
TuringTest
  • TuringTest
interesting...
anonymous
  • anonymous
That would make f(2) = 2?
anonymous
  • anonymous
When you plug it back in anyway...
asnaseer
  • asnaseer
yes
asnaseer
  • 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
asnaseer
  • asnaseer
we have two equations so far
asnaseer
  • asnaseer
maybe make use of f(0)?
asnaseer
  • asnaseer
which gives us:\[6\le c\le 10\]
maheshmeghwal9
  • maheshmeghwal9
\[x^2 - 4x + 6 ≤ (ax^2+bx+c)≤ 2x^2 - 8x +10\] \[\color{blue}{a(12)^2+b(12)+c}=182=\color{blue}{144a+12b+c.}\] \[\color{red}{a(17)^2+b(17)+c=?}\]\[6 \le c \le 10.\]
anonymous
  • anonymous
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?
asnaseer
  • asnaseer
f(1) is also interesting, it gives us:\[3\le a+b+c\le4\]
asnaseer
  • asnaseer
what do you mean by /strange/?
asnaseer
  • asnaseer
is there a particular approach that is used?
anonymous
  • anonymous
Well, the way they did it is a lot simpler and way different from this way...they completed the square...
asnaseer
  • asnaseer
hmmm... let me think over that approach...
maheshmeghwal9
  • maheshmeghwal9
Becoz we all are doing this type of question first time. & they had done it already. I really appreciate @asnaseer 's work:D
slaaibak
  • slaaibak
all three equations have a turning point (minimum) at x=2. so you could use that.
asnaseer
  • 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:\[a+b+c=4\tag{3b}\]
anonymous
  • anonymous
Does it become a geometric question when they complete the square? |dw:1343584198825:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
just thinking out loud, thought they might subtract squares then.
anonymous
  • anonymous
hmm no, that's not what I thought then.
anonymous
  • anonymous
I don't unserstand how they MUST have the same vertex though...
asnaseer
  • asnaseer
interesting observation @slaaibak - maybe we can use that to get the 3rd equation?
asnaseer
  • asnaseer
yes @Calcmathlete that ties up with @slaaibak's observation
slaaibak
  • 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
maheshmeghwal9
  • maheshmeghwal9
ya so they all have same vertex. as @Calcmathlete 's.............. said:D
anonymous
  • anonymous
Thin is, I don't understand how they MUST all have the same vertex. Is it because of how the parabolas work when graphed?
asnaseer
  • asnaseer
the LHS and RHS have the same vertex which fixes the vertex of the middle one
maheshmeghwal9
  • maheshmeghwal9
no becoz they all have same minimum value at x=2
maheshmeghwal9
  • maheshmeghwal9
thus they all have same vertex
slaaibak
  • 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
maheshmeghwal9
  • maheshmeghwal9
|dw:1343584535907:dw|
maheshmeghwal9
  • maheshmeghwal9
3 parabolas
maheshmeghwal9
  • maheshmeghwal9
3 functions quadratic......
asnaseer
  • asnaseer
very similar to the fact that x=2 gives the same value for LHS and RHS which fixes the value of f(2)
anonymous
  • anonymous
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?
slaaibak
  • 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
asnaseer
  • asnaseer
yes @Calcmathlete
slaaibak
  • slaaibak
yeah that's correct
anonymous
  • anonymous
Alright. Thank you! I wish I could give multiple medals...
asnaseer
  • asnaseer
yw :) and yes, I wish we had Olympic-style medals - Gold/Silver/Bronze :)
asnaseer
  • asnaseer
but then we would have to drug test everyone to make sure they were not cheating! :D
anonymous
  • anonymous
lol...
maheshmeghwal9
  • maheshmeghwal9
Graphs must be there.
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