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 points $\left(1,1\right)\mathbf{,}\left(2,0\right)$, such that ${\mathbf{\int }}_{\mathbf{1}}^{\mathbf{2}}\mathbf{f}\left(t\right)\mathbf{d}\mathbf{t}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

Find 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 $\left(1,1\right)$and $\left(2,0\right)$, such that${\int }_1^2\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}=-1$isdata-custom-editor="chemistry" $\mathrm{f}\left(\mathrm{t}\right)=20-28\mathrm{t}+9{\mathrm{t}}^2$.

Step-by-step solution

Consider the points and substitute these in the standard equation.

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

$$\begin{gathered} {\mathrm{f}}_1\left(\mathrm{t}\right)=\mathrm{a}+{\mathrm{bt}}_1+{\mathrm{ct}}_1^2 \\ \Rightarrow 1=\mathrm{a}+\mathrm{b}\left(1\right)+\mathrm{c}{\left(1\right)}^2 \\ {\mathrm{f}}_2\left(\mathrm{t}\right)=\mathrm{a}+{\mathrm{bt}}_2+{\mathrm{ct}}_2^2 \\ \Rightarrow 0=\mathrm{a}+\mathrm{b}\left(2\right)+\mathrm{c}{\left(2\right)}^2 \end{gathered}$$

Consider the integral equation of the polynomial $\mathrm{f}\left(\mathrm{t}\right)=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2$

$$\begin{gathered} {\int }_1^2\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}=-1 \\ \Rightarrow {\int }_1^2\left(\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2\right)d\mathrm{t}=-1 \\ \Rightarrow 2\mathrm{a}+2\mathrm{b}+\frac{8}{3}\mathrm{c}-\mathrm{a}-\frac{1}{2}\mathrm{b}-\frac{1}{3}\mathrm{c}=-1 \\ \therefore 6\mathrm{a}+9\mathrm{b}+14\mathrm{c}=-6 \end{gathered}$$

Rearrange the terms of the above equations.

Consider the simplified equations.

$$\begin{gathered} 1=\mathrm{a}+\mathrm{b}+\mathrm{c}....\left(1\right) \\ 0=\mathrm{a}+2\mathrm{b}+4\mathrm{c}....\left(2\right) \\ -6=6\mathrm{a}+9\mathrm{b}+14\mathrm{c}....\left(3\right) \end{gathered}$$

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

The values are

$$\begin{gathered} \left(\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 6& 9& 14\end{array}\right)\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\\ \mathrm{c}\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ -6\end{array}\right) \\ \mathrm{a}=20,\mathrm{b}=-28,\mathrm{c}=9 \end{gathered}$$

Substitute these values in the standard equation of the $\le 2$degree polynomial.

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)=\left(20\right)+\left(-28\right)\mathrm{t}+\left(9\right){\mathrm{t}}^2 \\ \therefore \mathrm{f}\left(\mathrm{t}\right)=20-28\mathrm{t}+9{\mathrm{t}}^2 \end{gathered}$$

Final answer

$\mathrm{f}\left(\mathrm{t}\right)=20-28\mathrm{t}+9{\mathrm{t}}^2$ is the equation of the polynomial of the degree$\le 2$ that passes through the points$\left(1,1\right)$ and$\left(2,0\right)$, such that ${\int }_1^2\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}=-1$