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I am writing this question as a self-answer, because I suspect, although I do not know for sure, what the OP (i.e. original poster) was trying to ask in this question which is now closed (make that now deleted).

I had intended to give a long-winded response comment, in an answer box, but the question was closed (and then deleted).

Further, the question that I think that the OP was trying to ask seems worthwhile.

So I will guess what the question was, present the question below, and then provide an answer. Note that I might be misinterpreting the OP's actual intent.

The Problem:

Find a polynomial $f(y)$ in the variable $y$ that satisfies the following $3$ constraints:

  • $f(1) = 12.$
  • $f(2) = 6.$
  • $f(3) = 18.$

Note: before I posted, I searched for duplicates, and found this question. However, that question didn't provide an explicit step-by-step solution. Therefore, I feel that this question is not precisely a duplicate.

David C. Ullrich
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user2661923
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2 Answers2

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You cannot find the polynomial that satisfies the given constraints, because if $f(x)$ is a solution, then, for any polynomial $g(x)$, the polynomial $$ f(x)+(x-1)(x-2)(x-3)g(x) $$ will satisfy the same constraints. But these are all the solutions! Indeed, if $f_1(x)$ is another solution, then the polynomial $$ p(x)=f_1(x)-f(x) $$ has the properties that $p(1)=p(2)=p(3)=0$ and so it is divisible by $x-1$, $x-2$ and $x-3$, hence by their product (use the remainder theorem) and therefore, writing $p(x)=(x-1)(x-2)(x-3)g(x)$, we get $$ f_1(x)=f(x)+(x-1)(x-2)(x-3)g(x) $$ as claimed.

Three conditions can determine at most three unknowns, so you can indeed look for solution of degree at most $2$. Using the system you do in your answer is good, but you can also directly use a Lagrange interpolating polynomial \begin{align} f(x) &= 12\frac{(x-2)(x-3)}{(1-2)(1-3)} + 6\dfrac{(x-1)(x-3)}{(2-1)(2-3)} + 18\frac{(x-1)(x-2)}{(3-1)(3-2)} \\[6px] &=6(x^2-5x+6)-6(x^2-4x+3)+9(x^2-3x+2) \\[12px] &=9x^2-33x+36 \end{align} Can you see why the polynomial satisfies the constraints? If you evaluate it at $1$, the second and third summand become zero, whereas the first summand is precisely $6$. Similarly when evaluating at $2$ and $3$.

egreg
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  • Yes, I agree that the answer that I gave is not unique, and I have edited my answer accordingly. However, from the perspective of preCalculus students (who never met LaGrange) I feel that it is valuable to keep things simple. The thrust of my answer is problem solving, with enough editing to keep any assertions that I make valid. – user2661923 Sep 05 '21 at 16:32
  • But these are all solutions! I didn't got this line , please explain – Lalit Tolani Sep 05 '21 at 16:35
  • @LalitTolani When you have found one polynomial $f(x)$ satisfying the constraints, you can prove that if $f_1(x)$ also satisfies the constraints, then $f_1(x)=f(x)+(x-1)(x-2)(x-3)g(x)$, for some polynomial $g(x)$; the converse is easy. – egreg Sep 05 '21 at 21:07
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Find a polynomial $f(y)$ in the variable $y$ that satisfies the following $3$ constraints:

  • $f(1) = 12.$
  • $f(2) = 6.$
  • $f(3) = 18.$

Edit
Per the answer of egreg, there is no unique solution. However, as indicated below, if you arbitrarily decide that $f(y)$ must be a 2nd degree polynomial, then the answer is unique.

Just as two constraints similar to the above may be defined by a 1st degree polynomial, three such constraints may be defined by a 2nd degree polynomial. In fact $n$ such constraints may be defined by a polynomial of degree $(n-1)$.

This means that the simplest solution is to assume that $f(y)$ is a 2nd degree polynomial. Under this assumption, real numbers $a,b,c$ must be chosen so that $f(y) = ay^2 + by + c.$ This leads to the following three equations in three unknowns:

$a + b + c = 12~:~~~~~$ based on $y = 1$.
$4a + 2b + c = 6~:~~~~~$ based on $y = 2$.
$9a + 3b + c = 18~:~~~~~$ based on $y = 3$.

Subtracting the 1st equation from the 2nd gives
$3a + b = -6$.

Subtracting the 1st equation from the 3rd gives
$8a + 2b = 6 \implies 4a + b = 3.$

Subtracting the 1st derived equation from the 2nd derived equation gives
$a = 9.$

Plugging this back into the 1st derived equation gives
$b = -33.$

Plugging these $a,b$ values into the 1st original equation gives
$c = 36.$

Therefore $f(y) = 9y^2 - 33y + 36.$

user2661923
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  • Why are you assuming it to be a quadratic polynomial? – Lalit Tolani Sep 05 '21 at 16:03
  • @LalitTolani Nice catch, I have edited my answer accordingly. That is, the simplest solution is to assume that $f(y)$ is a 2nd degree polynomial. – user2661923 Sep 05 '21 at 16:05
  • What if it is a cubic polynomial of type $ax^3+bx^2+cx$ or $ax^3+bx^2+d$ or $ax^3+cx+d$ – Lalit Tolani Sep 05 '21 at 16:08
  • @LalitTolani Certainly possible, and in fact the polynomial could be of any degree $n \geq 2$, with all coefficients set to $0$, for the terms $x^k ~: k < (n-2)$. However, in the preCalculus world, providing a solution in a 2nd degree equation will, I think, be the easiest to understand. – user2661923 Sep 05 '21 at 16:10
  • @LalitTolani Also, I want the original poster, and other preCalculus readers to explicitly see $3$ equations in $3$ unknowns, leading to a unique answer. The assumption that $f(y)$ is a 2nd degree polynomial serves this purpose. – user2661923 Sep 05 '21 at 16:13