Question

select the inequality that represents the graph.

$y - 1 < \frac{1}{2}\left(x + 3\
ight)$

$y - 2 > \frac{1}{2}\left(x + 1\
ight)$

$y + 1 > \frac{1}{2}\left(x - 3\
ight)$

$y + 2 < \frac{1}{2}\left(x - 1\
ight)$

Explanation:

Step1: Find the slope and a point on the line

The line passes through points \((-6, 0)\) and \((0, 3)\). The slope \(m=\frac{3 - 0}{0 - (-6)}=\frac{3}{6}=\frac{1}{2}\). Using point - slope form \(y - y_1=m(x - x_1)\), using the point \((-6,0)\), we have \(y-0=\frac{1}{2}(x + 6)\), or using the point \((0,3)\), \(y - 3=\frac{1}{2}(x-0)\). Also, we can check the given options. Let's check the first option: \(y - 1=\frac{1}{2}(x + 3)\). If we rewrite it in slope - intercept form \(y=\frac{1}{2}x+\frac{3}{2}+1=\frac{1}{2}x+\frac{5}{2}\). The second option: \(y - 2=\frac{1}{2}(x + 1)\) can be rewritten as \(y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{5}{2}\)? Wait, no, \(y - 2=\frac{1}{2}(x + 1)\) gives \(y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{5}{2}\)? Wait, let's find two points from the line in the graph. The line crosses the \(y\) - axis at \(y = 3\)? Wait, no, looking at the graph, the line passes through \((-6,0)\) and \((0,3)\)? Wait, no, when \(x = 0\), the \(y\) - intercept seems to be \(y = 3\)? Wait, no, the first option: \(y-1=\frac{1}{2}(x + 3)\), when \(x=-3\), \(y = 1\). The second option: \(y - 2=\frac{1}{2}(x + 1)\), when \(x=-1\), \(y = 2\). The third option: \(y + 1=\frac{1}{2}(x - 3)\), when \(x = 3\), \(y=-1\). The fourth option: \(y + 2=\frac{1}{2}(x - 1)\), when \(x = 1\), \(y=-2\).

Wait, let's check the slope. The slope of the line in the graph: from the points \((-6,0)\) and \((0,3)\), slope \(m=\frac{3-0}{0 + 6}=\frac{1}{2}\). Now, let's check the inequality sign. The region is shaded above the line, and the line is dashed? Wait, no, the line in the graph is a dashed line? Wait, the first option: \(y - 1=\frac{1}{2}(x + 3)\), let's rewrite it as \(y=\frac{1}{2}x+\frac{3}{2}+1=\frac{1}{2}x+\frac{5}{2}\). The second option: \(y - 2=\frac{1}{2}(x + 1)\) gives \(y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{5}{2}\)? Wait, no, \(y - 2=\frac{1}{2}(x + 1)\) is \(y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{5}{2}\)? Wait, no, \(\frac{1}{2}(x + 1)=\frac{1}{2}x+\frac{1}{2}\), then \(y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{5}{2}\). Wait, the line in the graph: when \(x=-6\), \(y = 0\). Let's plug \(x=-6\) into the first option: \(y-1=\frac{1}{2}(-6 + 3)=\frac{1}{2}(-3)=-\frac{3}{2}\), so \(y=1-\frac{3}{2}=-\frac{1}{2} eq0\). Second option: \(y - 2=\frac{1}{2}(-6 + 1)=\frac{1}{2}(-5)=-\frac{5}{2}\), \(y=2-\frac{5}{2}=-\frac{1}{2} eq0\). Third option: \(y + 1=\frac{1}{2}(-6-3)=\frac{1}{2}(-9)=-\frac{9}{2}\), \(y=-1-\frac{9}{2}=-\frac{11}{2} eq0\). Fourth option: \(y + 2=\frac{1}{2}(-6 - 1)=\frac{1}{2}(-7)=-\frac{7}{2}\), \(y=-2-\frac{7}{2}=-\frac{11}{2} eq0\). Wait, maybe I made a mistake. Let's check the \(y\) - intercept. The line in the graph: when \(x = 0\), what's \(y\)? Looking at the graph, the shaded region is above the line, and the line passes through \((-6,0)\) and \((0,3)\)? Wait, no, the first option: \(y-1=\frac{1}{2}(x + 3)\), when \(x = 0\), \(y-1=\frac{3}{2}\), \(y=\frac{5}{2}=2.5\). The second option: \(y - 2=\frac{1}{2}(x + 1)\), when \(x = 0\), \(y-2=\frac{1}{2}\), \(y=\frac{5}{2}=2.5\). Wait, maybe the line is \(y-2=\frac{1}{2}(x + 1)\)? Wait, no, let's check the points. Let's take a point in the shaded region, say \((0,3)\). Plug into the first option: \(3-1=\frac{1}{2}(0 + 3)\) → \(2=\frac{3}{2}\)? No. Second option: \(3 - 2=\frac{1}{2}(0+1)\) → \(1=\frac{1}{2}\)? No. Wait, maybe the slope is \(\frac{1}{2}\), and the line passes through \((-6,0)\) and \((0,3)\). The equation of the line is \(y=\frac{1}{2}x + 3\). Let's rewrite the first option: \(y-1=\frac{1}{2}(x + 3)\) → \(y=\frac{1}{2}x+\frac{3}{2}+1=\frac{1}{2}x+\frac{5}{2}=2.5\). Not 3. Second option: \(y - 2=\frac{1}{2}(x + 1)\) → \(y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{5}{2}=2.5\). Third option: \(y + 1=\frac{1}{2}(x - 3)\) → \(y=\frac{1}{2}x-\frac{3}{2}-1=\frac{1}{2}x-\frac{5}{2}=-2.5\). Fourth option: \(y + 2=\frac{1}{2}(x - 1)\) → \(y=\frac{1}{2}x-\frac{1}{2}-2=\frac{1}{2}x-\frac{5}{2}=-2.5\). Wait, maybe I misread the graph. Let's look at the graph again. The line passes through \((-6,0)\) and \((0,3)\), so the slope is \(\frac{3-0}{0 - (-6)}=\frac{1}{2}\), and the equation is \(y=\frac{1}{2}x + 3\). Now, let's check the inequality. The shaded region is above the line, and the line is dashed (since the inequality is strict? Wait, the options have \(<\) or \(>\)? Wait, the first option is \(y - 1<\frac{1}{2}(x + 3)\)? No, the user's options: let's re - check the options. The first option: \(y - 1<\frac{1}{2}(x + 3)\)? Wait, no, the user's options:

1. \(y - 1<\frac{1}{2}(x + 3)\)
2. \(y - 2>\frac{1}{2}(x + 1)\)
3. \(y + 1>\frac{1}{2}(x - 3)\)
4. \(y + 2<\frac{1}{2}(x - 1)\)

Wait, maybe the line is \(y - 2=\frac{1}{2}(x + 1)\) and the inequality is \(y-2>\frac{1}{2}(x + 1)\) (since the shaded region is above the line). Let's take a point in the shaded region, say \((0,3)\). Plug into \(y - 2>\frac{1}{2}(x + 1)\): \(3 - 2>\frac{1}{2}(0 + 1)\) → \(1>\frac{1}{2}\), which is true. Let's check another point, say \((-1,2)\): \(2 - 2>\frac{1}{2}(-1 + 1)\) → \(0>0\)? No, wait, \((-1,2)\) is on the line. The line is dashed, so the inequality is strict. Take \((0,4)\): \(4 - 2>\frac{1}{2}(0 + 1)\) → \(2>\frac{1}{2}\), true. Take \((-6,0)\): \(0 - 2>\frac{1}{2}(-6 + 1)\) → \(-2>\frac{1}{2}(-5)\) → \(-2>-\frac{5}{2}\) (since \(-\frac{5}{2}=-2.5\) and \(-2>-2.5\)), which is true.

Now, let's check the first option: \(y - 1<\frac{1}{2}(x + 3)\). Take \((0,3)\): \(3 - 1<\frac{1}{2}(0 + 3)\) → \(2<\frac{3}{2}\), false. Third option: \(y + 1>\frac{1}{2}(x - 3)\). Take \((0,3)\): \(3 + 1>\frac{1}{2}(0 - 3)\) → \(4>-\frac{3}{2}\), true, but let's check the slope. The slope of \(y + 1=\frac{1}{2}(x - 3)\) is \(\frac{1}{2}\), but the \(y\) - intercept is \(y=\frac{1}{2}x-\frac{3}{2}-1=\frac{1}{2}x-\frac{5}{2}\), which is not the same as our line. Fourth option: \(y + 2<\frac{1}{2}(x - 1)\). Take \((0,3)\): \(3 + 2<\frac{1}{2}(0 - 1)\) → \(5<-\frac{1}{2}\), false.

So the correct inequality is \(y - 2>\frac{1}{2}(x + 1)\)

Answer:

\(y - 2>\frac{1}{2}(x + 1)\) (assuming the second option is \(y - 2>\frac{1}{2}(x + 1)\) as the original options might have a typo in the symbol, but based on the graph and the point - checking, this is the correct one)