Question

Find all the polynomials of degree$\mathbf{\le }\mathbf{2}$[a polynomial of the form $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{t}\mathbf{+}\mathit{c}{\mathit{t}}^{\mathbf{4}}$] whose graph goes through the points role="math" localid="1659337996890" $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$and$\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}$, such that${\mathit{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

Short answer

The polynomial of degree $\le 2$, whose graph goes through the points$(1,1)$ and$(3,3)$ such that$f^{\text{'}}\left(2\right)=1$ is $f\left(t\right)=3c+\left(1-4c\right)t+c{t}^2$.

Step-by-step solution

Consider the points and substitute these in the standard equation.

A polynomial of degree 2 is of the form$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{t}\mathbf{+}\mathit{c}{\mathit{t}}^{\mathbf{4}}$. Consider a polynomial of degree 2 and substitute given point in them as:

$$\begin{gathered} f_1\left(t\right)=a+b{t}_1+c{t}_1^2 \\ 1=a+b\left(1\right)+c{\left(1\right)}^2 \end{gathered}$$

$$\begin{gathered} f_2\left(t\right)=a+b{t}_2+c{t}_2^2 \\ 3=a+b\left(3\right)+c(3{)}^2 \end{gathered}$$

Consider the derivative of the polynomial $f\left(t\right)=a+bt+c{t}^2$as $f^{\text{'}}\left(2\right)=1$:

$$\begin{gathered} f^{\text{'}}\left(t\right)=b+2ct \\ {f}^{\text{'}}\left(2\right)=b+2c\left(2\right) \\ 1=b+4c \end{gathered}$$

Rearrange the terms of the above equations

Consider the simplified equations.

$$\begin{gathered} 1=a+b+c......\left(1\right) \\ 3=a+3b+9c......\left(2\right) \\ 1=0+b+4c......\left(3\right) \end{gathered}$$

Solve the above equations (1), (2) and (3).

Arrange the equations (1), (2) and (3) into matrix form:

$\left(\begin{array}{ccc}1& 1& 1\\ 1& 3& 9\\ 0& 1& 4\end{array}\right)\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}1\\ 3\\ 1\end{array}\right)$

Upon solving the values of$a,b$ and c are obtained as$a=3c,b=1-4c,c=c$

Substitute these values in the standard equation of the $\le 2$degree polynomial.role="math" localid="1659338786441" $$\begin{gathered} f\left(t\right)=\left(3c\right)+(1-4c)t+\left(c\right)t^2 \\ f\left(t\right)=3c+(1-4c)t+c{t}^2 \end{gathered}$$

The polynomial of degree $\le 2$, whose graph goes through the points $(1,1)$and $(3,3)$such that$f^{\text{'}}\left(2\right)=1$ is $f\left(t\right)=3c+\left(1-4c\right)t+c{t}^2$.