91影视

The equation is, $f\left(t\right)=a+bt+c{t}^2+d{t}^3+e{t}^4$

Substitute the points in the above equation to form the equations.

$$\begin{gathered} \left(1,1\right): \\ f\left(1\right)=a+b\left(1\right)+c{\left(1\right)}^2+d{\left(1\right)}^3+e{\left(1\right)}^4 \\ =1 \\ a+b+c+d+e=1鈥�鈥�\left(1\right) \end{gathered}$$

$$\begin{gathered} \begin{array}{l}(2,-1):\end{array} \\ f\left(2\right)=a+b\left(2\right)+c{\left(2\right)}^2+d{\left(2\right)}^3+e{\left(2\right)}^4 \\ =1 \end{gathered}$$

$a+2b+4c+8d+16e=-1鈥�鈥�\left(2\right)$

$$\begin{gathered} \left(3,-59\right) \\ f\left(3\right)=a+b\left(3\right)+c{\left(3\right)}^2+d{\left(3\right)}^3+e{\left(3\right)}^4 \\ =-59 \end{gathered}$$

$$\begin{gathered} \left(-1,5\right): \\ f\left(-1\right)=a+b\left(-1\right)+c{\left(-1\right)}^2+d{\left(-1\right)}^3+e{\left(-1\right)}^4 \\ =5 \end{gathered}$$

$$\begin{gathered} \left(-2,-29\right): \\ f\left(-2\right)=a+b\left(-2\right)+c{\left(-2\right)}^2+d{\left(-2\right)}^3+e{\left(-2\right)}^4 \\ =-29 \end{gathered}$$

Re-write the equations in terms of a matrix.

$\left|\begin{array}{c}a+b+c+d+e=1\\ a+2b+4c+8d+16e=-1\\ a+3b+9c+27d+81e=-59\\ a-b+c-d+e=5\\ a-2b+4c-8d+16e=-29\end{array}\right|$

The matrix form is,

$\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 2& 4& 8& 16& -1\\ 1& 3& 9& 27& 81& -59\\ 1& -1& 1& -1& 1& -1\\ 1& -2& 4& -8& 16& -29\end{array}\right]$

Consider the matrix.

$\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 2& 4& 8& 16& -1\\ 1& 3& 9& 27& 81& -59\\ 1& -1& 1& -1& 1& -1\\ 1& -2& 4& -8& 16& -29\end{array}\right]$

Using row Echelon form to reduce the matrix.

$$\begin{gathered} \left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 2& 4& 8& 16& -1\\ 1& 3& 9& 27& 81& -59\\ 1& -1& 1& -1& 1& -1\\ 1& -2& 4& -8& 16& -29\end{array}\right]=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 0& -3& 3& -9& 15& -30\\ 0& 0& 10& 20& 90& -80\\ 0& 0& 0& 8& 8& 2\\ 0& 0& 0& 0& -12& 21\end{array}\right] \\ =\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0& -5\\ 0& 0& 1& 0& 0& 4\\ 0& 0& 0& 1& 0& 3\\ 0& 0& 0& 0& 1& -2\end{array}\right] \end{gathered}$$

The values are, $a=1,b=-5,c=4,d=3,e=-2$.

Therefore, the polynomial is,

$$\begin{gathered} f\left(t\right)=a+bt+c{t}^2+d{t}^3+e{t}^4 \\ =1-5t+4t^2+3t^2-2t^4 \end{gathered}$$

The graphical representation is,