Question

1. Determine the polynomial function whose graph passes through the points (0, 10), (1, 7), (3, -11), and (4, -14). Be sure to include a sketch of the polynomial functions, showing the points. Solve using the Gauss-Jordan method or Gaussian elimination with back substitution. Show the matrix and rovw operation used for each step. 2. The figure below shows the flow of traffic (in vehicles per hour) through a network of streets. 300 200 100 500 YA Y 600 400 450 350 l 600 400 What is the system of equation that models this network? Solve this system of equations. HINT: You may use your calculator and the RREF shortcut. Find the traffic flow when (a) (b) (c) = 285 and x2 = 625.

Answer 1

The polynomial equation can be written as

ax⁴+bx³+cx²+dx+e =0

Here,we have to find the value of a,b,c,d and e.

The matrix equation

$\begin{bmatrix} 0 & 0& 0& 0& 1& 10\\ 1& 1& 1 & 1& 1&7 \\ 81 & 27 & 9& 3& 1 &-11 \\ 256 & 64& 16 & 4& 1& -14 \end{bmatrix}$

Swapping the 1st column and swap the second column and 1st row,we have

$\begin{bmatrix} 1 & 1& 1& 1& 1& 7\\ 0& 0& 0 & 0& 1&10 \\ 81 & 27 & 9& 3& 1 &-11 \\ 256 & 64& 16 & 4& 1& -14 \end{bmatrix}$

Multiply the first row by 81 and substract the first row from the third and restore it,we get

$\begin{bmatrix} 1 & 1& 1& 1& 1& 7\\ 0& 0& 0 & 0& 1&10 \\ 0& -54 & -72& -78& -80 &-578 \\ 256 & 64& 16 & 4& 1& -14 \end{bmatrix}$

Again multiply the first row by 256 and substract it from 4th row

$\begin{bmatrix} 1 & 1& 1& 1& 1& 7\\ 0& 0& 0 & 0& 1&10 \\ 0& -54 & -72& -78& -80 &-578 \\ 0 & -192& -240 & -252& -255& -1806 \end{bmatrix}$

Divide the third row by -54 and swap the third and second row and Subtract the second row from 1st-row

$\begin{bmatrix} 1 & 0& -1/3& -4/9& -13/27& -100/27\\ 0& 1& 4/3& 13/9& 40/27&289/27 \\ 0& -54 & -72& -78& -80 &-578 \\ 0 & -192& -240 & -252& -255& -1806 \end{bmatrix}$

c.Multiply the second row by -192, Subtract the second row from the 4th and restore it

$\begin{bmatrix} 1 & 0& -1/3& -4/9& -13/27& -100/27\\ 0& 1& 4/3& 13/9& 40/27&289/27 \\ 0& -54 & -72& -78& -80 &-578 \\ 0 & 0& 16 & 76/3& 265/9& 2242/9 \end{bmatrix}$

.Make the pivot in third column by dividing the 4th row by 16 and swap the 4th and 3rd rows and Multiply the third row by -1/3

$\begin{bmatrix} 1 & 0& -1/3& -4/9& -13/27& -100/27\\ 0& 1& 4/3& 13/9& 40/27&289/27 \\ 0& 0 & -1/3& -19/36& -265/432 &-1121/216 \\ 0 & 0& 16 & 76/3& 265/9& 2242/9 \end{bmatrix}$

Subtract the 3rd from the 1st and multiply the third by 4/3

$\begin{bmatrix} 1 & 0& 0& 1/12& 19/144& 107/72\\ 0& 1& 4/3& 13/9& 40/27&289/27 \\ 0& 0 & 4/3& 19/9& 265/108 &1121/54 \\ 0 & 0&0 & 0& 1& 10 \end{bmatrix}$

Subtract the thrird from the second and multiply the 4th row by 19/44

$\begin{bmatrix} 1 & 0& 0& 1/12& 19/144& 107/72\\ 0& 1& 0& -2/3& -35/36&-181/18 \\ 0& 0 & 1& 19/12& 265/144 &1121/72 \\ 0 & 0&0 & 0& 19/144& 95/72 \end{bmatrix}$

Subtract the 4th from first, multiply by -35/36 and again subtract the 4th from the second we get

$\begin{bmatrix} 1 & 0& 0& 1/12&0& 1/6\\ 0& 1& 0& -2/3& 0&-1/3 \\ 0& 0 & 1& 19/12& 265/144 &1121/72 \\ 0 & 0&0 & 0& 1& 10 \end{bmatrix}$

Finally, multiply the 4th row by 265/144 and subtract the 4th from the 3rd

$\begin{bmatrix} 1 & 0& 0& 1/12&0& 1/6\\ 0& 1& 0& -2/3& 0&-1/3 \\ 0& 0 & 1& 19/12& 0 &-17/6 \\ 0 & 0&0 & 0& 1& 10 \end{bmatrix}$

Thus the equation is

a =1/6 -d/12

b= -1/3+d/4

c=17/6-19d/12

e=10 and d can be any number.

Considering d=0 ,we have

a= 1/6,b=-1/3 ,c=17/6,e=10

Equation is

x⁴/6 -x³/3+17x²/6+10 =0

Sketch