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Assume that $P$ is a quadratic polynomial such that $P(1+x) = P(3-x)$, and $P(x)\geq 1$. If $P(0) = 13$, how could we find $P(1)$?

If $P(0) = 13$, then $P$ is of the form $P(x) = ax^2+bx+13$. Since $P(1+x) = P(3-x)$, $P(4) = P(0) =13$ and therefore, $16a+4b = 0$, $b = -4a$. And if $P(-\frac{b}{2a}) = 1$, then $4a-8a + 13 = 1$, leading us to $a = 3$ and $P(x) = 3x^2-12x+13$. From which we could conclude that $P(1) = 4$.

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    Hint: what is $P'(2)$? – Greg Martin Jan 16 '23 at 19:28
  • Apologies for that! It's obvious that $\frac{dP}{dx}|_{x = 2} = 4a+b = 0$, yes. –  Jan 16 '23 at 19:43
  • Suggestion: Rather than editing the question when you found the answer yourself, you can post an answer on your question yourself, and it is a better method as it would avoid any confusions to those seeing this question in the future. – mrtechtroid Jan 17 '23 at 17:14
  • Sure, I'll keep that in mind. –  Jan 17 '23 at 17:51

1 Answers1

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Hint…consider the minimum point$$P(-\frac{b}{2a})=1$$

David Quinn
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    It's not clear whether the problem description is meant to imply that the lower bound $1$ is actually attained. – Greg Martin Jan 16 '23 at 20:32
  • @GregMartin well it does say $P(x)\geq 1$ and not $P(x)>1$ so I think we should infer that the minimum value of $P(x)$ is $1$ – David Quinn Jan 16 '23 at 21:01
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    That's not what the notation $\ge$ means.... – Greg Martin Jan 16 '23 at 21:38
  • @GregMartin yes I know, and the question statement is not precise. But I think we would have to assume that that is what was intended otherwise the problem is insoluble. – David Quinn Jan 17 '23 at 09:34