which linear inequality is represented by the graph?\n$y < 3x + 2$\n$y > 3x + 2$\n$y < \\frac{1}{3}x +…

Explanation:

Step1: Find the slope.

The slope $m$ of a line passing through two points $(x_1,y_1)=(0,2)$ and $(x_2,y_2)=(-3,-7)$ is given by the formula $m=\frac{y_2 - y_1}{x_2 - x_1}$. So, $m=\frac{-7 - 2}{-3-0}=\frac{-9}{-3}=3$.

Step2: Determine the y - intercept.

The line crosses the y - axis at the point $(0,2)$, so the y - intercept $b = 2$. The equation of the line in slope - intercept form is $y=mx + b$, which is $y = 3x+2$.

Step3: Determine the inequality.

The line is dashed, so the inequality is either $y>3x + 2$ or $y<3x + 2$. We can test a point in the shaded region, say $(0,0)$. Substitute $x = 0$ and $y = 0$ into the inequalities. For $y<3x + 2$, we have $0<3\times0+2$ (i.e., $0 < 2$) which is true. For $y>3x + 2$, we have $0>3\times0+2$ (i.e., $0>2$) which is false.

Answer:

$y<3x + 2$