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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)

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

we have two equations so far

maybe make use of f(0)?

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

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?

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

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

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

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...

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

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

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

|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)

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.