Let $P(x)$ be the interpolating polynomial for the data $(0, 0)$, $(0.5, y)$, $(1, 3)$ and $(2, 2)$. Find $y$ if the coefficient of $x^3$ in $P(x)$ is $6$.
I tried finding the Lagrange interpolating polynomial in terms of $y$, and then basically making the coefficients of each piece add up to $6$ and then solve for $y$. This however did not seem to be correct when I verified my work. Is this the correct method of solving?
I got $30.22$ as my answer but this doesn't seem right and when I plug it back in I get $75.2593$ on the $x^3$ term.
The Lagrange Interpolating polynomial is $P(x)=\frac{(x-0.5)(x-1)(x-2)}{(0-0.5)(0-1)(0-2)}*0+\frac{(x-0)(x-1)(x-2)}{(0.5-0)(0.5-1)(0.5-2)}*y+\frac{x(x-0.5)(x-2)}{(1-0)(1-0.5)(1-2)}*3+\frac{x(x-0.5)(x-1)}{(2-0)(2-0.5)(2-1)}*2$
Bring it to standard normal form for polynomials (meaning $a_3x^3+a_2x^2+a_1x+1_0$) and find $y$ such that $a_3=6$.
The answer: To save work, notice that we are not interested in the entire polynomial, but just the coefficients of $x^3$. The first fraction in the polynomial is zero. The second one, the coefficient of $x^3$ will be $\frac{8}{3}y$. In the third fraction, we will have $6x^3$, and in the fourth fraction we will have $\frac{2}{3}x^3$. So overall, the coefficient of $x^3$ in $P$ is $\frac{8}{3}y+6+\frac{2}{3}$. You want this to be equal to $6$, this means that $y=-\frac{1}{4}$