Write an inequality that best represents the graph.

The graph is given in this question.

The given graph is:

From the given graph, it can be noticed that the boundary is a line that is passing through the points $\left(0,1\right)$and $\left(\frac{3}{2},0\right)$

Therefore, the equation of the boundary can be found out by finding the equation of the line which is passing through the points $\left(0,1\right)$and $\left(\frac{3}{2},0\right)$

It is known that the equation of a line passing through the points $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$is:

Therefore, the equation of the line passing through the points $\left(0,1\right)$and $\left(\frac{3}{2},0\right)$is:

role="math" localid="1647750169358" $\begin{array}{c}y-1=\frac{0-1}{\frac{3}{2}-0}\left(x-0\right)\\ y-1=\frac{-1}{\frac{3}{2}}x\\ y-1=\frac{-2}{3}x\\ y=-\frac{2}{3}x+1\end{array}$

Therefore, the equation of the boundary line is $y=-\frac{2}{3}x+1$

The equation of the boundary line is $y=-\frac{2}{3}x+1$

From the given graph, it can be noticed that the boundary is drawn by the solid line that implies the inequality will contain the equality sign.

Therefore, the inequality shown in the graph can be either -r $y\le -\frac{2}{3}x+1$or $y\ge -\frac{2}{3}x+1$

Take any point which lies in the shaded region of the given graph and substitute that point in the inequalities $y\le -\frac{2}{3}x+1$and $y\ge -\frac{2}{3}x+1$The equation which gets satisfied after substituting that point is the inequality shown in the graph.

From the given graph, it can be noticed that the point$(0,0)$lies in the shaded region.

Substitute the point $(0,0)$in the inequality $y\le -\frac{2}{3}x+1$

$\begin{array}{c}y\le -\frac{2}{3}x+1\\ 0\le \frac{-2}{3}\left(0\right)+1\\ 0\le 0+1\\ 0\le 1\end{array}$

Substitute the point $(0,0)$in the inequality $y\ge -\frac{2}{3}x+1$

$\begin{array}{c}y\ge -\frac{2}{3}x+1\\ 0\ge \frac{-2}{3}\left(0\right)+1\\ 0\ge 0+1\\ 0\ge 1\end{array}$

Therefore, it can be noticed that point $(0,0)$satisfies the inequality $y\le -\frac{2}{3}x+1$and point $(0,0)$does not satisfy the inequality $y\ge -\frac{2}{3}x+1$

Therefore, the inequality shown in the graph is$y\le -\frac{2}{3}x+1$