Determine a polynomial function whose graph passes through the given points. 5: (0,-5); (1,-4) ; (-1,-9); (2,-3),

Answer:The interpolating polynomial is[tex]p(x) = \frac{1}{2}x^3-\frac{3}{2}+2x-5[/tex].Step-by-step explanation:First, notice that we want to calculate the interpolating polynomial through the points (0,-5); (1,-4) ; (-1,-9); (2,-3). This means that we want to find a polynomial [tex]p(x)[/tex] such that[tex]p(-1) = -9[/tex], [tex]p(0) = -5[/tex] [tex]p(1) = -4[/tex] and [tex]p(2) = -3[/tex].We will have four equations, so our polynomial will be, at most, of degree 3. Let us write[tex]p(x) = a_0 + a_1x +a_2x^2 +a_3x^3.[/tex]The coordinates give us the following equations:[tex] \begin{cases} -9= p(-1) &= a_0 - a_1 +a_2 -a_3 \\ -5=p(0) &= a_0 \\-4=p(1) &= a_0 + a_1 +a_2 +a_3 \\ -3 = p(2) &= a_0 + 2a_1 +4a_2 +8a_3 \end{cases}[/tex]Notice that from the second equation we know that [tex]a_0 =-5[/tex]. Then, we obtained the linear system of equations[tex] \begin{cases} -9 &= -5 - a_1 +a_2 -a_3 \\-4 &= -5 + a_1 +a_2 +a_3 \\ -3 &= -5 + 2a_1 +4a_2 +8a_3 \end{cases}[/tex]which is equivalent to[tex]\begin{cases} - a_1 +a_2 -a_3 &= -4\\ a_1 +a_2 +a_3 &= 1\\ 2a_1 +4a_2 +8a_3 &= 2\end{cases}[/tex].So, we have reduced our interpolation problem to solve a linear system of equations. Now, notice that if we add the first two equations of the system we obtain[tex]2a_2=-3[/tex] that yields [tex]a_2 = -\frac{3}{2}[/tex].Then, our system becomes[tex]\begin{cases}-a_1 -\frac{3}{2}-a_3 &= -4\\a_1 -\frac{3}{2} +a_3 &= 1\\ 2a_1 -6 +8a_3 &= 2\end{cases}[/tex]which is equivalent to[tex]\begin{cases}-a_1-a_3 &=-\frac{5}{2}\\a_1+a_3 &=\frac{5}{2}\\ 2a_1+8a_3 &=8\end{cases}[/tex].Recall that now the first two equations are just the same, so we will use the first and third ones:[tex]\begin{cases}-a_1-a_3 &=-\frac{5}{2}\\2a_1+8a_3 &= 8\end{cases}[/tex].If we multiply the first one and add it to the second we get:[tex]6a_3 = 3[/tex] that yields [tex]a_3 = \frac{1}{2}[/tex].Thus, substituting this value in the first equation:[tex]-a_1-\frac{1}{2} = -\frac{5}{2}[/tex] which is equivalent to [tex]-a_1=-\frac{4}{2}[/tex]. Then, [tex]a_1=2[/tex].Summing up all our results we get that the interpolating polynomial is[tex]p(x) = \frac{1}{2}x^3-\frac{3}{2}+2x-5[/tex].