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).

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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
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my first thought is to try factoring and see if that takes us somewhere
Aren't they both unfactorable?
I guess, I'm just brainstorming (and multitasking)

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