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The slope of${\mathbf{x}}_{\mathbf{2}}$from the equation that can be seen is -3. A negative value indicates that${\mathbf{x}}_{\mathbf{2}}$ has an inverse relation with y and a higher value denotes that its of high magnitude.

Given, $\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{+}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}$

Now to plot this equation, make a table

Given, $\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{+}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$

$\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)}{\mathbf{x}}_{\mathbf{2}}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}\mathbf{4}{\mathbf{x}}_{\mathbf{2}}$

Now to plot this equation, make a table

Given, $\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{+}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{2}$

$\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{(}\mathbf{2}\mathbf{)}{\mathbf{x}}_{\mathbf{2}}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{-}\mathbf{5}{\mathbf{x}}_{\mathbf{2}}$

Now to plot this equation, make a table

As it can be seen in the graph, for the value of${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}$, E(y) passes through (1, -1)${\mathbf{x}}_{\mathbf{2}}$ whenis$\mathbf{1}\mathbf{猢�}{\mathbf{x}}_{\mathbf{2}}\mathbf{猢�}\mathbf{3}$ . And for every change in the value of${\mathbf{x}}_{\mathbf{1}}$out slope of the line changes and the line becomes steeper.

For the given value of ${\mathbf{x}}_{\mathbf{2}}$between $\mathbf{1}\mathbf{猢�}{\mathbf{x}}_{\mathbf{2}}\mathbf{猢�}\mathbf{3}$, the changes in the value of ${\mathbf{x}}_{\mathbf{1}}$makes the slope of the line becomes steeper as the slope parameter increases from 3 to 4 to 5 for the values of ${\mathbf{x}}_{\mathbf{1}}$as 0, 1, and 2.

E(y) changes by 1 to 17 to units when the value of ${\mathbf{x}}_{\mathbf{2}}$is $\mathbf{1}\mathbf{猢�}{\mathbf{x}}_{\mathbf{2}}\mathbf{猢�}\mathbf{3}$and ${\mathbf{x}}_1$is $\mathbf{3}\mathbf{猢�}{\mathbf{x}}_{\mathbf{1}}\mathbf{猢�}\mathbf{5}$

Given, $\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{+}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}$for${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{3}$ and${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1}$

$\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$for${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}$

Given, role="math" localid="1649834551696" $\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{+}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{5}$and${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{3}$

$\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{)}\mathbf{-}\mathbf{3}\mathbf{(}\mathbf{3}\mathbf{)}\mathbf{-}\mathbf{5}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$for ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

$\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{17}$