Question

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

Explanation:

Step1: Find the slope of the line

The line passes through \((0, 2)\) and \((-3, -7)\). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-7 - 2}{-3 - 0}=\frac{-9}{-3}=3\). So the equation of the line is \(y = 3x + 2\) (using slope - intercept form \(y=mx + b\), where \(b = 2\) from the point \((0,2)\)).

Step2: Determine the inequality symbol

The line is dashed (so the inequality is strict, either \(>\) or \(<\)), and the shaded region is to the left of the line. Let's test a point in the shaded region, say \((0,0)\) (wait, no, the shaded region is on the left. Let's take \((-3,0)\) which is in the shaded region. Plug into \(y\) and \(3x + 2\): For \(x=-3\), \(3x + 2=3\times(-3)+2=-9 + 2=-7\). The \(y\) - value of the point \((-3,0)\) is \(0\), and \(0>-7\). Wait, no, let's check the direction of the shading. The shaded region is below or above? Wait, the line goes through \((0,2)\) and \((-3,-7)\). Let's take the origin \((0,0)\): \(0\) vs \(3\times0 + 2=2\). \(0<2\)? But the shaded region includes \((-3, -7)\)? Wait, no, the shaded region is on the left side of the line. Wait, when \(x = 0\), the line is at \(y = 2\). The shaded region is below the line? Wait, no, let's check the slope again. Wait, the two points are \((0,2)\) and \((-3,-7)\). The slope is \(\frac{-7 - 2}{-3-0}=\frac{-9}{-3}=3\), correct. Now, the line is dashed, so the inequality is either \(y>3x + 2\) or \(y<3x + 2\). Let's take a point in the shaded area, say \((-3,0)\). Plug into \(y\) and \(3x + 2\): \(y = 0\), \(3x+2=3\times(-3)+2=-7\). Since \(0>-7\), and the shaded region is above the line? Wait, no, when \(x = 0\), the line is at \(y = 2\). The shaded region is to the left of the line. Let's take \(x=-1\), then \(3x + 2=3\times(-1)+2=-1\). A point in the shaded region with \(x=-1\) would have \(y\) - value. Let's see, the line at \(x=-1\) is \(y=-1\). The shaded region: if we take \(y = 0\) at \(x=-1\), \(0>-1\), so \(y>3x + 2\) would satisfy \(0>3\times(-1)+2=-1\). Wait, but let's check the options. The options are \(y<3x + 2\), \(y>3x + 2\), \(y<\frac{1}{3}x + 2\), \(y>\frac{1}{3}x + 2\). Wait, I made a mistake in the slope earlier? Wait, no, \((0,2)\) and \((-3,-7)\): the slope is \(\frac{-7 - 2}{-3-0}=\frac{-9}{-3}=3\), correct. Wait, maybe I mixed up the direction. Wait, the line is increasing (positive slope). The shaded region is on the side where \(y\) is less than \(3x + 2\)? Wait, no, let's take the point \((0,0)\): \(0\) vs \(3\times0+2 = 2\). \(0<2\), but is \((0,0)\) in the shaded region? Looking at the graph, the shaded region is on the left of the line, which includes \((-3,-7)\) and \((-5,0)\) etc. Wait, maybe I messed up the slope calculation. Wait, \((0,2)\) and \((-3,-7)\): the run is \(-3-0=-3\), the rise is \(-7 - 2=-9\), so slope is \(\frac{-9}{-3}=3\), correct. Now, the line is dashed, so the inequality is strict. Now, let's check the shading: the shaded region is below the line? No, when \(x = 0\), the line is at \(y = 2\), and the shaded region is on the left, which is where \(y\) values are less than \(3x + 2\)? Wait, no, let's take \(x=-3\), \(3x + 2=-7\), and a point in the shaded region with \(x=-3\) has \(y\) - value greater than \(-7\) (like \(y = 0\) at \(x=-3\) is in the shaded region? Wait, the point \((-3,0)\): \(0>-7\), so \(y>3x + 2\) when \(x=-3\). Wait, but when \(x = 0\), \(3x+2 = 2\), and the shaded region at \(x = 0\) is below \(y = 2\)? No, the graph shows the shaded region is on the left, including points like \((-3,-7)\) (the dot) and going up. Wait, maybe the correct inequality is \(y<3x + 2\)? Wait, no, let's re - examine. The line has a slope of \(3\), equation \(y = 3x+2\). The shaded region: if we pick a point in the shaded area, say \((-3,-8)\) (below the line at \(x=-3\), since the line at \(x=-3\) is \(y=-7\)). Then \(y=-8\), and \(3x + 2=-7\). So \(-8<-7\), which would mean \(y<3x + 2\). Ah, I see, I made a mistake in choosing the test point earlier. The point \((-3,-8)\) is in the shaded region (since the shaded region goes down to \((-3,-7)\) and below? Wait, the dot is at \((-3,-7)\), and the shaded region is below the line? Wait, the line is dashed, so the inequality is \(y<3x + 2\) because the shaded region is below the dashed line.

Answer:

\(y<3x + 2\) (the first option: \(y<3x + 2\))