Let \(\alpha, \beta, \gamma, \delta\) be real numbers. We ask when it is possible to find a polynomial \(f\) over \(\RR\), of degree not more than 2, such that \(f(-1) = \alpha\), \(f(1) = \beta\), \(f(3) = \gamma\) and \(f(0) = \delta\). Prove that this is possible if and only if \[ 3\alpha + 6\beta - \gamma - 8\delta = 0. \]
Solution
Write \[ f(x)=ax^2+bx+c. \] Since \(f(0)=\delta\), we have \(c=\delta\). The equations at \(-1\) and \(1\) give \[ a-b+\delta=\alpha,\qquad a+b+\delta=\beta, \] so \[ a=\frac{\alpha+\beta-2\delta}{2}, \qquad b=\frac{\beta-\alpha}{2}. \] Then \[ \begin{aligned} f(3) &=9a+3b+\delta\\ &=3\alpha+6\beta-8\delta. \end{aligned} \] Thus the condition \(f(3)=\gamma\) is exactly \[ 3\alpha+6\beta-\gamma-8\delta=0. \] This proves necessity. Conversely, if this condition holds, the displayed values of \(a,b,c\) define a polynomial of degree at most \(2\) satisfying all four interpolation conditions. Hence the condition is also sufficient.