Question

Find the polynomial of degree 3 [a polynomial of the form$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{t}\mathbf{+}\mathit{c}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathit{d}{\mathit{t}}^{\mathbf{3}}$] whose graph goes through the points $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$and $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{15}\mathbf{)}$. Sketch the graph of this cubic.

Short answer

The graphical representation of the cubic polynomial$f\left(t\right)=1+2t-1t^2-2t^3$is,

Step-by-step solution

Consider the points and form the equations

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

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

$$\begin{gathered} \left(0,1\right): \\ F\left(0\right)=a+b\left(0\right)+c{\left(0\right)}^2+d{\left(0\right)}^3 \\ =1 \\ a=1......\left(1\right) \end{gathered}$$

$$\begin{gathered} \left(1,0\right): \\ f\left(1\right)=a+b\left(1\right)+c{\left(1\right)}^2+d{\left(1\right)}^3 \\ =0 \\ a+b+c+d=0.......\left(2\right) \end{gathered}$$

$$\begin{gathered} \left(-1,0\right): \\ f\left(-1\right)=a+b\left(-1\right)+c{\left(-1\right)}^2+d{\left(-1\right)}^3 \\ a-b+c-d=0......\left(3\right) \end{gathered}$$

$$\begin{gathered} \left(2,-15\right): \\ f\left(2\right)=a+b\left(2\right)+c{\left(2\right)}^2+d{\left(2\right)}^3 \\ a+2b+4c+8d=-15.......\left(4\right) \end{gathered}$$

Consider the matrix

Re-write the equations in terms of a matrix.

$\left|\begin{array}{c}a=1\\ a+b+c+d=0\\ a-b+c-d=0\\ a+2b+4c+8d=-15\end{array}\right|$

The matrix form is,

$\left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 1& 1& 1& 1& 0\\ 1& -1& 1& -1& 0\\ 1& 2& 4& 8& -15\end{array}\right]$

Solve the matrix

Consider the matrix.

$\left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 1& 1& 1& 1& 0\\ 1& -1& 1& -1& 0\\ 1& 2& 4& 8& -15\end{array}\right]$

Using row Echelon form to reduce the matrix.

$$\begin{gathered} \left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 1& 1& 1& 1& 0\\ 1& -1& 1& -1& 0\\ 1& 2& 4& 8& -15\end{array}\right]=\left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 0& 2& 4& 8& -16\\ 0& 0& 3& 3& -9\\ 0& 0& 0& -2& 4\end{array}\right] \\ =\left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 0& 1& 0& 0& 2\\ 0& 0& 1& 0& -1\\ 0& 0& 0& -1& -2\end{array}\right] \end{gathered}$$

The values are,$a=1,b=2,c=-1,d=-2$

Therefore, the polynomial is,

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

Sketch the graph.

The graphical representation is,