91Ó°ÊÓ

The slope of$x_2$from the equation that can be seen is -3. A negative value indicates that$x_2$ has an inverse relation with y and a higher value denotes that its of high magnitude.

Given

$$\begin{gathered} E\left(y\right)=2+x_1-3x_2-x_1{x}_{2}for{x}_1=0 \\ =2+0-3x_2-0×x_2 \\ =2-3x_2 \end{gathered}$$

Now to plot this equation, make a table

Given

$$\begin{gathered} E\left(y\right)=2+x_1-3x_2-x_1{x}_{2}for{x}_1=1 \\ =2+1-3x_2-1×x_2 \\ =3-4x_2 \end{gathered}$$

Now to plot this equation, make a table

Given

$$\begin{gathered} E\left(y\right)=2+x_1-3x_2-x_1{x}_{2}for{x}_1=2 \\ =2+2-3x_2-2×x_2 \\ =4-5x_2 \end{gathered}$$

Now to plot this equation, make a table

As it can be seen in the graph, for the value of$x_1=0,1,2$ , E(y) passes through (1, -1) when$x_2$ is. $1≤x_2≤3$And for every change in the value of$x_1$out slope of the line changes and the line becomes steeper.

For the given value of $x_2$between $1≤x_2≤3$, the changes in the value of $x_1$makes the slope of the line becomes steeper as the slope parameter increases from 3 to 4 to 5 for the values of $x_1$as 0, 1, and 2.

E(y) changes by 1 to 17 to units when the value of $x_2$is $1≤x_2≤3$and $x_1$is $3≤x_1≤5$

Given,

role="math" localid="1649798013525" $$\begin{gathered} E\left(y\right)=2+x_1-3x_2-x_1{x}_{2}for{x}_1=3and{x}_2=1 \\ y=2+3-3×1-3×1 \\ y=-1 \end{gathered}$$

Given,

$$\begin{gathered} E\left(y\right)=2+x_1-3x_2-x_1{x}_{2}for{x}_1=5and{x}_2=3 \\ y=2+5-3×3-5×3 \\ y=-17 \end{gathered}$$