Question

Find the polynomial of degree $\mathbf{\le }\mathbf{2}$[a polynomial of the form $\mathbf{f}\left(t\right)\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{bt}\mathbf{+}{\mathbf{ct}}^{\mathbf{2}}$] whose graph goes through the pointsdata-custom-editor="chemistry" $\left(1,p\right)\mathbf{,}\left(2,q\right)\mathbf{,}\left(3,r\right)$ wheredata-custom-editor="chemistry" $\mathbf{p}\mathbf{,}\mathbf{q}\mathbf{,}\mathbf{r}$ are arbitary constants. Does such a polynomial exists for all values of data-custom-editor="chemistry" $\mathbf{p}\mathbf{,}\mathbf{q}\mathbf{,}\mathbf{r}$?

Short answer

The polynomial of degree $\le 2$[a polynomial of the form data-custom-editor="chemistry" $\mathrm{f}\left(\mathrm{t}\right)=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2$] whose graph goes through the points data-custom-editor="chemistry" $\left(1,\mathrm{p}\right),\left(2,\mathrm{q}\right),\left(3,\mathrm{r}\right)$exists for all values ofdata-custom-editor="chemistry" $\mathrm{p},\mathrm{q},\mathrm{r}$ as long as determinant of this system does not equal to zero.

Step-by-step solution

Conider the points and substitute these in the standard equations.

The polynomial of degree 2 is $\mathrm{f}\left(\mathrm{t}\right)=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2$

$$\begin{gathered} \mathrm{p}=\mathrm{a}+\mathrm{b}\left(1\right)+\mathrm{c}{\left(1\right)}^2 \\ \mathrm{q}=\mathrm{a}+\mathrm{b}\left(2\right)+\mathrm{c}{\left(2\right)}^2 \\ \mathrm{r}=\mathrm{a}+\mathrm{b}\left(3\right)+\mathrm{c}{\left(3\right)}^2 \end{gathered}$$

Rearrange the terms of the above equations.

Consider the simplified equations.

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

The matrix form of the bove equations.

Represent the above equations in terms of matrix

$\left(\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 1& 3& 9\end{array}\right)\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\\ \mathrm{c}\end{array}\right)=\left(\begin{array}{c}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right)$

Final answer

For all the values of$\mathrm{p},\mathrm{q},\mathrm{r}$ polynomial exists as long as the determinant of the matrix is not equal to zero.