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

The derivative of the equation is,$f\text{'}\left(t\right)=b+2ct+3d{t}^2$

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

localid="1659349415086" $$\begin{gathered} (1,1): \\ f\left(1\right)=a+b\left(1\right)+c{\left(1\right)}^2+d{\left(1\right)}^3 \\ =1 \\ a+b+c+d=1....\left(1\right) \\ (2,5): \\ f\left(2\right)=a+b\left(2\right)+c{\left(2\right)}^2+d{\left(2\right)}^3 \\ =-1 \\ a+2b+4c+8d=5......\left(2\right) \\ (1,2): \\ f\text{'}\left(1\right)=b\left(3\right)+2\left(c\right)+3d{\left(3\right)}^2 \\ =2 \\ b+2c+3d=2.......\left(3\right) \\ (2,9): \\ f\text{'}\left(2\right)=b\left(3\right)+2c{\left(3\right)}^2+3d{\left(3\right)}^2 \\ =9 \\ b+4c+12d=9.............\left(4\right) \end{gathered}$$

Re-write the equations in terms of a matrix.

$\left|\begin{array}{c}a+b+c+d=1\\ a+2b+4c+8d=5\\ b+2c+3d=2\\ b+4c+12d=9\end{array}\right|$

The matrix form is,

$\left|\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 4& 8& 5\\ 0& 1& 2& 3& 2\\ 0& 1& 4& 12& 9\end{array}\right|$

Consider the matrix.

$\left|\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 4& 8& 5\\ 0& 1& 2& 3& 2\\ 0& 1& 4& 12& 9\end{array}\right|$

Using row Echelon form to reduce the matrix.

$$\begin{gathered} \left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 4& 8& 5\\ 0& 1& 2& 3& 2\\ 0& 1& 4& 12& 9\end{array}\right]=\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 3& 7& 4\\ 0& 0& -1& -4& -2\\ 0& 0& 0& 1& 3\end{array}\right] \\ =\left[\begin{array}{ccccc}1& 0& 0& 0& -5\\ 0& 1& 0& 0& 13\\ 0& 0& 1& 0& -10\\ 0& 0& 0& 1& 3\end{array}\right] \end{gathered}$$

The values are, $a=-5,b=13,c=-10,d=3$

Therefore, the polynomial is,

$$\begin{gathered} f\left(t\right)=a+bt+c{t}^2+d{t}^3 \\ =-5+13t-10t^2+3t^3 \end{gathered}$$

The graphical representation of $f\left(t\right)=-5+13t-10t^2+3t^3$is, ss