Question

this is the graph of a linear inequality. write the inequality in slope - intercept form.

Explanation:

Step1: Find the slope-intercept form of the line

First, identify two points on the line. From the graph, we can see that the line passes through \((0, 1)\) and \((1, 0)\). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - 1}{1 - 0}=- 1\). The \(y\)-intercept \(b\) is \(1\) (since the line crosses the \(y\)-axis at \((0,1)\)). So the equation of the line in slope - intercept form (\(y = mx + b\)) is \(y=-x + 1\).

Step2: Determine the inequality symbol

The line is solid (since the boundary is included in the solution set, we can tell from the graph's line style, although in the given graph it looks like a solid line? Wait, looking at the graph, the line is a solid line? Wait, no, wait the shading: the shaded region. Let's check the direction of the inequality. Let's pick a test point, say \((0,0)\). Plug into \(y=-x + 1\), we get \(0\)? \(0\) vs \(-0 + 1=1\). \(0<1\), but is \((0,0)\) in the shaded region? Wait, looking at the graph, the shaded region: when \(x = 0\), \(y\) values less than \(1\) and greater than \(1\)? Wait, no, let's re - examine the graph. Wait, the line goes from \((-5,8)\) (wait, no, earlier points: wait, when \(x=-5\), \(y = 8\)? Wait, maybe my initial points were wrong. Let's take two clear points. Let's see, when \(x = 0\), \(y=1\); when \(x = 1\), \(y = 0\); when \(x=2\), \(y=-1\); when \(x=-1\), \(y = 2\). So the slope is still \(-1\), \(y\)-intercept \(1\). Now, the shaded region: let's take a point in the shaded region, say \((-5,8)\). Plug into \(y=-x + 1\): \(8\) vs \(-(-5)+1=5 + 1=6\). \(8>6\). So the inequality is \(y\geq - x+1\)? Wait, no, wait \((-5,8)\): \(y = 8\), \(-x + 1=5 + 1 = 6\), \(8>6\). Wait, but when \(x = 0\), the \(y\)-intercept is \(1\). Wait, maybe the line is \(y=-x + 1\), and the shaded region is above the line? Wait, no, let's check the shading again. Wait, the graph's shaded area: when \(x\) decreases (goes to the left), \(y\) increases, and when \(x\) increases, \(y\) decreases. Let's take the point \((0,0)\): is \((0,0)\) in the shaded region? Looking at the graph, the shaded region at \(x = 0\) includes \(y\) values from, say, below \(1\) and above \(1\)? No, wait, the line is \(y=-x + 1\). Let's check the test point \((-2,3)\). Plug into \(y=-x + 1\): \(3\) vs \(-(-2)+1=2 + 1=3\). So \((-2,3)\) is on the line? Wait, no, \(-2\) plugged into \(y=-x + 1\) gives \(y = 3\), so \((-2,3)\) is on the line. Wait, maybe the line is \(y=-x + 1\) and the inequality is \(y\geq - x + 1\)? Wait, no, let's re - do the slope. Let's take two points: \((x_1,y_1)=(-5,8)\) and \((x_2,y_2)=(3,-2)\). Then slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-2 - 8}{3-(-5)}=\frac{-10}{8}=-\frac{5}{4}\)? Wait, I think I made a mistake earlier. Wait, the graph: let's count the grid. Each grid is 1 unit. So from \((0,1)\) to \((1,0)\): that's a slope of \(-1\). From \((0,1)\) to \((-1,2)\): slope is \(\frac{2 - 1}{-1-0}=-1\). From \((0,1)\) to \((1,0)\): slope \(-1\). Now, the boundary line: is it solid or dashed? The problem says "linear inequality", if the line is solid, the inequality is \(\geq\) or \(\leq\); if dashed, \(\gt\) or \(\lt\). Looking at the graph, the line appears to be solid (since the boundary is included). Now, let's take a test point in the shaded region. Let's take \((-5,8)\): \(y = 8\), \( - x+1=5 + 1 = 6\), \(8>6\). So \(y\geq - x + 1\)? Wait, no, \(8>6\), so \(y\geq - x + 1\) would include \(y = 8\) when \(x=-5\). Wait, but when \(x = 0\), \(y = 1\) is on the line. Let's check another point: \((0,2)\). Plug into \(y=-x + 1\): \(2\) vs \(1\), \(2>1\). Is \((0,2)\) in the shaded region? Looking at the graph, at \(x = 0\), the shaded region goes up to \(y = 8\) (at \(x=-5\)) and down to \(y=-8\) (at \(x = 8\))? Wait, maybe I messed up the direction. Wait, the correct way: the slope - intercept form of the line is \(y=-x + 1\). The line is solid, so the inequality is either \(y\geq - x + 1\) or \(y\leq - x + 1\). Let's take the point \((0,0)\): \(0\) vs \(1\), \(0<1\). Is \((0,0)\) in the shaded region? Looking at the graph, at \(x = 0\), the shaded region includes \(y\) values less than \(1\) (below the line) and greater than \(1\) (above the line)? No, that can't be. Wait, the graph's shaded area: it's a triangle - like shape? Wait, no, the graph shows a shaded region that is above the line \(y=-x + 1\)? Wait, no, when \(x\) increases, \(y\) decreases. Let's take \(x=-5\), \(y = 8\): \(8=-(-5)+1=6\)? No, \(8 eq6\). Wait, I think my initial calculation of the line is wrong. Let's recalculate the line. Let's find two points on the line. From the graph, the line passes through \((0,1)\) and \((1,0)\), so slope \(m=\frac{0 - 1}{1 - 0}=-1\), equation \(y=-x + 1\). Now, the shaded region: let's check the point \((-2,3)\). Plug into \(y=-x + 1\): \(3=-(-2)+1=3\), so \((-2,3)\) is on the line. The point \((-3,4)\): \(4=-(-3)+1=4\), also on the line. The point \((-4,5)\): \(5=-(-4)+1=5\), on the line. The point \((-5,6)\): \(6=-(-5)+1=6\), on the line. Wait, earlier I thought \((-5,8)\) was on the line, but that's a mistake. So the line is \(y=-x + 1\), and the shaded region: when \(x=-5\), the \(y\)-value on the line is \(6\), but the shaded region goes up to \(y = 8\) at \(x=-5\). So \(8>6\), which means \(y\geq - x + 1\). Wait, but when \(x = 0\), \(y = 1\) is on the line, and the shaded region includes above the line? Wait, no, let's take a point below the line, say \((0,0)\): \(0<1\), is \((0,0)\) in the shaded region? Looking at the graph, at \(x = 0\), the shaded region includes \(y\) values from, like, below \(1\) (since at \(x = 0\), the line is at \(y = 1\), and the shaded region goes down to \(y=-8\) at \(x = 8\)). Wait, I'm confused. Wait, maybe the line is \(y=-x + 1\), and the inequality is \(y\leq - x + 1\)? Wait, no, let's check the test point \((0,0)\): \(0\leq - 0+1=1\), which is true. Is \((0,0)\) in the shaded region? Looking at the graph, yes, \((0,0)\) is in the shaded region (the blue area). Wait, earlier mistake: when I took \((-5,8)\), that's not on the line. The line is \(y=-x + 1\), so at \(x=-5\), \(y = 6\), but the shaded region at \(x=-5\) is above \(y = 6\) (up to \(y = 8\)) and below \(y = 6\)? No, the graph's shaded area: it's a region that is below the line \(y=-x + 1\) when \(x\) is positive and above when \(x\) is negative? No, that can't be. Wait, the correct approach: the slope - intercept form of a linear inequality is \(y>mx + b\), \(y\geq mx + b\), \(y<mx + b\), or \(y\leq mx + b\). The line is solid, so we use \(\geq\) or \(\leq\). Let's take the test point \((0,0)\). Plug into \(y=-x + 1\): \(0\leq1\), which is true. And \((0,0)\) is in the shaded region. Now, take the point \((-1,2)\): \(2=-(-1)+1=2\), so \((-1,2)\) is on the line. Take the point \((-2,3)\): \(3=-(-2)+1=3\), on the line. So the line is \(y=-x + 1\), and the shaded region includes all points where \(y\leq - x + 1\)? Wait, no, \((0,0)\) gives \(0\leq1\), which is true, and \((0,0)\) is in the shaded region. But when \(x=-5\), the \(y\)-value on the line is \(6\), and the shaded region at \(x=-5\) has \(y = 8\), which is \(8>6\), so \(8\leq6\) is false. So my test point selection was wrong. Let's take a point in the shaded region that is not on the line. Let's take \((-2,4)\). Plug into \(y=-x + 1\): \(4=-(-2)+1=3\), \(4>3\). So \(y>-x + 1\)? But the line is solid, so it should be \(y\geq - x + 1\). Ah, because the line is solid, the boundary is included, so the inequality is \(y\geq - x + 1\)? Wait, \((-2,4)\): \(4\geq3\), which is true. \((0,0)\): \(0\geq1\)? No, \(0<1\), so \((0,0)\) is not in the region defined by \(y\geq - x + 1\). So there's a contradiction. Wait, I must have messed up the line equation. Let's re - calculate the line. Let's find two points on the line. From the graph, the line passes through \((0,1)\) and \((1,0)\), so slope \(m=-1\), equation \(y=-x + 1\). Now, the shaded region: let's look at the direction of the shading. The line is going from the top - left to the bottom - right. The shaded region is above the line? Wait, when \(x\) decreases (moves to the left), \(y\) increases. So for a given \(x\), the shaded region has \(y\) values greater than the line's \(y\)-value. Let's take \(x = 0\), line \(y = 1\), shaded region at \(x = 0\) includes \(y>1\) (since at \(x = 0\), the shaded area goes up to \(y = 8\) or something? Wait, no, the graph's shaded area: when \(x=-5\), the \(y\)-coordinate of the top of the shaded region is \(8\), and the line at \(x=-5\) is \(y=-(-5)+1=6\), so \(8>6\). When \(x = 5\), the line at \(x = 5\) is \(y=-5 + 1=-4\), and the shaded region at \(x = 5\) has \(y>-4\) (since the bottom of the shaded region at \(x = 5\) is, say, \(y=-8\)? No, the graph's bottom at \(x = 8\) is \(y=-8\), and the line at \(x = 8\) is \(y=-8 + 1=-7\), so \(-8<-7\), which would mean \(y\leq - x + 1\) at \(x = 8\). This is confusing. Wait, maybe the correct line is \(y=-x + 1\) and the inequality is \(y\geq - x + 1\) because the line is solid and the shaded region is above the line. Wait, let's check the standard method:

1. Find the equation of the boundary line: \(y=-x + 1\) (slope \(-1\), \(y\)-intercept \(1\)).
2. Determine if the line is solid (included) or dashed (not included). The line in the graph appears to be solid (so we use \(\geq\) or \(\leq\)).
3. Choose a test point not on the line. Let's use \((0,2)\). Plug into \(y=-x + 1\): \(2\) vs \(-0 + 1 = 1\). \(2>1\), and \((0,2)\) is in the shaded region. So the inequality is \(y\geq - x + 1\).

Answer:

\(y\geq -x + 1\)