Question

Find the polynomial of degree 4 whose graph goes through the points (−3,−296),(−2,−60),(−2,−60), (0,4), (2,4) and (3,−110).

Answer 1

Answer on Question #61885 – Math – Linear Algebra

Question

Find the polynomial of degree 4 whose graph goes through the points $(-3, -296), (-2, -60), (-2, -60), (0, 4), (2, 4)$ and $(3, -110)$.

Solution

The general form of the polynomial of degree 4 is

$$y = f(x) = a_1 x^4 + a_2 x^3 + a_3 x^2 + a_4 x + a_5.$$

Table 1

Substituting values from Table 1 into (1) obtain the system as follows

$$\left\{ \begin{array}{l} a_1 \cdot (-3)^4 + a_2 \cdot (-3)^3 + a_3 \cdot (-3)^2 + a_4 \cdot (-3) + a_5 = -296, \\ a_1 \cdot (-2)^4 + a_2 \cdot (-2)^3 + a_3 \cdot (-2)^2 + a_4 \cdot (-2) + a_5 = -260, \\ \quad a_1 \cdot (0)^4 + a_2 \cdot (0)^3 + a_3 \cdot (0)^2 + a_4 \cdot (0) + a_5 = 4, \\ \quad a_1 \cdot (2)^4 + a_2 \cdot (2)^3 + a_3 \cdot (2)^2 + a_4 \cdot (2) + a_5 = 4, \\ \quad a_1 \cdot (3)^4 + a_2 \cdot (3)^3 + a_3 \cdot (3)^2 + a_4 \cdot (3) + a_5 = -110, \end{array} \right.$$ $$\left\{ \begin{array}{l} 81a_1 - 27a_2 + 9a_3 - 3a_4 + a_5 = -296, \\ 16a_1 - 8a_2 + 4a_3 - 2a_4 + a_5 = -260, \\ 0 \cdot a_1 + 0 \cdot a_2 + 0 \cdot a_3 + 0 \cdot a_4 + a_5 = 4, \\ \quad 16a_1 + 8a_2 + 4a_3 + 2a_4 + a_5 = 4, \\ 81a_1 + 27a_2 + 9a_3 + 3a_4 + a_5 = -110, \end{array} \right.$$ $$\left( \begin{array}{rrrrr} 81 & -27 & 9 & -3 & 1 \\ 16 & -8 & 4 & -2 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ 16 & 8 & 4 & 2 & 1 \\ 81 & 27 & 9 & 3 & 1 \end{array} \right) \left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right) = \left( \begin{array}{c} -296 \\ -260 \\ 4 \\ 4 \\ -110 \end{array} \right)$$

Let $C = \left( \begin{array}{rrrrr}

81 & -27 & 9 & -3 & 1 \\

16 & -8 & 4 & -2 & 1 \\

0 & 0 & 0 & 0 & 1 \\

16 & 8 & 4 & 2 & 1 \\

81 & 27 & 9 & 3 & 1

\end{array} \right)$.

Then the following matrix can be computed by means of the function MINVERSE from Microsoft Excel:

$$C^{-1} = \frac{1}{360} \left( \begin{array}{cccccc} 4 & -9 & 10 & -9 & 4 \\ -12 & 18 & 0 & -18 & 12 \\ -16 & 81 & -130 & 81 & -16 \\ 48 & -162 & 0 & 162 & -48 \\ 0 & 0 & 360 & 0 & 0 \end{array} \right) = \left( \begin{array}{cccccc} \frac{1}{90} & -\frac{1}{40} & \frac{1}{36} & -\frac{1}{40} & \frac{1}{90} \\ -\frac{1}{30} & \frac{1}{20} & 0 & -\frac{1}{20} & \frac{1}{30} \\ -\frac{2}{45} & \frac{9}{40} & -\frac{13}{36} & \frac{9}{40} & -\frac{2}{45} \\ \frac{2}{15} & -\frac{9}{20} & 0 & \frac{9}{20} & -\frac{2}{15} \\ 0 & 0 & 1 & 0 & 0 \end{array} \right) \approx$$ $$\approx \left( \begin{array}{cccccc} 0.011 & -0.025 & 0.028 & -0.025 & 0.011 \\ -0.033 & 0.05 & 0 & -0.05 & 0.033 \\ -0.044 & 0.225 & -0.361 & 0.225 & -0.044 \\ 0.133 & -0.45 & 0 & 0.45 & -0.133 \\ 0 & 0 & 1 & 0 & 0 \end{array} \right).$$

Identities

$$C^{-1}C = E = \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right), \quad (3)$$ $$E \left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right) = \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right) = \left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right) \quad (4)$$

hold true.

Multiplying (2) on the left by $C^{-1}$ and using (3), (4) and function MMULT from Microsoft Excel obtain

$$\begin{array}{l} \left( \begin{array}{cccccc} \frac{1}{90} & -\frac{1}{40} & \frac{1}{36} & -\frac{1}{40} & \frac{1}{90} \\ -\frac{1}{30} & \frac{1}{20} & 0 & -\frac{1}{20} & \frac{1}{30} \\ -\frac{2}{45} & \frac{9}{40} & -\frac{13}{36} & \frac{9}{40} & -\frac{2}{45} \\ \frac{2}{15} & -\frac{9}{20} & 0 & \frac{9}{20} & -\frac{2}{15} \\ 0 & 0 & 1 & 0 & 0 \end{array} \right) \left( \begin{array}{cccccc} 81 & -27 & 9 & -3 & 1 \\ 16 & -8 & 4 & -2 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ 16 & 8 & 4 & 2 & 1 \\ 81 & 27 & 9 & 3 & 1 \end{array} \right) \left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right) = \\ = \left( \begin{array}{cccccc} \frac{1}{90} & -\frac{1}{40} & \frac{1}{36} & -\frac{1}{40} & \frac{1}{90} \\ -\frac{1}{30} & \frac{1}{20} & 0 & -\frac{1}{20} & \frac{1}{30} \\ -\frac{2}{45} & \frac{9}{40} & -\frac{13}{36} & \frac{9}{40} & -\frac{2}{45} \\ \frac{2}{15} & -\frac{9}{20} & 0 & \frac{9}{20} & -\frac{2}{15} \\ 0 & 0 & 1 & 0 & 0 \end{array} \right) \left( \begin{array}{c} -296 \\ -260 \\ 4 \\ 4 \\ -110 \end{array} \right), \end{array}$$ $$\left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{array} \right) = \left( \begin{array}{c} 2 \\ -7 \\ -41 \\ 94 \\ 4 \end{array} \right).$$

Thus, the polynomial of degree 4 whose graph goes through the points $(-3, -296)$, $(-2, -60)$, $(-2, -60)$, $(0, 4)$, $(2, 4)$ and $(3, -110)$ will be $2x^4 - 7x^3 - 41x^2 + 94x + 4$.

Answer: $2x^4 - 7x^3 - 41x^2 + 94x + 4$.

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