Question

find the inequality represented by the graph.

Explanation:

Step1: Find two points on the line

From the graph, we can identify two points on the line: \((-5, -5)\) and \((-3, -3)\) (wait, actually looking at the axes, let's re - check. Wait, the x - axis and y - axis: let's take two points. Let's see, when \(x = 0\), what's \(y\)? Wait, maybe the points are \((-5, -5)\) and \((0, - 2)\)? Wait, no, let's calculate the slope. Let's take two points: \((-5,-5)\) and \((0, - 2)\)? Wait, no, looking at the graph, the line passes through \((-5,-5)\) and \((5,0)\)? Wait, maybe I misread. Wait, let's do it properly. Let's find two points with integer coordinates. Let's see, when \(x=-5\), \(y = - 5\); when \(x = 0\), \(y=-2\)? No, wait, let's calculate the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points: \((-5,-5)\) and \((5,0)\). Then \(m=\frac{0 - (-5)}{5-(-5)}=\frac{5}{10}=\frac{1}{2}\). Wait, no, maybe the points are \((-5,-5)\) and \((0, - 2)\)? Wait, slope \(m=\frac{-2-(-5)}{0 - (-5)}=\frac{3}{5}\)? No, this is getting confusing. Wait, another approach: the line equation. Let's assume the line is in the form \(y=mx + b\). Let's find two points. From the graph, when \(x = 0\), \(y=-2\) (y - intercept \(b=-2\)). And when \(y = 0\), \(x = 5\)? Wait, no, let's check the points. Wait, the blue line: let's take two points with integer coordinates. Let's say \((-5,-5)\) and \((5,0)\) are on the line? Wait, no, let's calculate the slope between \((-5,-5)\) and \((0, - 2)\): \(m=\frac{-2-(-5)}{0 - (-5)}=\frac{3}{5}\)? No, maybe the correct points are \((-5,-5)\) and \((0, - 2)\) is wrong. Wait, let's look at the graph again. The line passes through \((-5,-5)\) and \((5,0)\)? Wait, when \(x=-5\), \(y=-5\); when \(x = 5\), \(y = 0\). Then the slope \(m=\frac{0 - (-5)}{5-(-5)}=\frac{5}{10}=\frac{1}{2}\). Then the equation of the line is \(y=\frac{1}{2}x + b\). Plug in \((5,0)\): \(0=\frac{1}{2}(5)+b\), \(b=-\frac{5}{2}\). No, that doesn't seem right. Wait, maybe the line is \(y=\frac{1}{2}x-2\)? Wait, when \(x = 0\), \(y=-2\); when \(x = 4\), \(y = 0\)? No, this is not working. Wait, let's start over.

Step1: Identify the line equation

First, find two points on the boundary line. From the graph, the boundary line passes through \((-5, - 5)\) and \((0,-2)\)? Wait, no, let's check the coordinates. Wait, the x - axis and y - axis: the horizontal axis is y - axis? Wait, no, the labels: the vertical axis is x - axis (downward) and horizontal is y - axis (rightward). Wait, that's a bit confusing. Wait, the standard is x - axis horizontal (left - right) and y - axis vertical (up - down). But in the graph, the vertical axis is labeled x (downward) and horizontal is y (rightward). So we need to adjust. So the coordinates: a point \((x,y)\) where x is the vertical coordinate (downward) and y is the horizontal coordinate (rightward). So let's re - define: let's take the vertical axis as x (so moving down increases x) and horizontal as y (moving right increases y). So the two points on the line: when x = 5 (down 5 units), y=-5 (left 5 units)? No, this is too confusing. Wait, maybe the graph is a bit mislabeled, and we can consider the standard x (horizontal) and y (vertical) with the labels swapped. Let's assume that the horizontal axis is x and vertical is y (ignoring the labels for a moment). Then the line passes through \((-5,-5)\) and \((5,0)\). Wait, no, the shaded region is above the line? Wait, the blue region is above the line. Let's find the slope between \((-5,-5)\) and \((5,0)\): \(m=\frac{0 - (-5)}{5-(-5)}=\frac{5}{10}=\frac{1}{2}\). The equation of the line is \(y=\frac{1}{2}x + b\). Plug in \((5,0)\): \(0=\frac{1}{2}(5)+b\Rightarrow b =-\frac{5}{2}\). No, that's not correct. Wait, another way: let's take two points \((0,-2)\) (y - intercept) and \((5,0)\). Then slope \(m=\frac{0 - (-2)}{5-0}=\frac{2}{5}\). Then the equation is \(y=\frac{2}{5}x-2\). Wait, when \(x = 0\), \(y=-2\); when \(x = 5\), \(y=\frac{2}{5}(5)-2=2 - 2 = 0\). Yes, that works. Now, the line is solid (since the boundary is a solid line), so the inequality will include equality. The shaded region is above the line, so we use \(y\geq\) the line equation. Wait, but let's check the points. Wait, if the line is \(y=\frac{2}{5}x-2\), and the shaded region is above the line, then the inequality is \(y\geq\frac{2}{5}x-2\). But let's verify with a point in the shaded region. Let's take \((0,0)\): plug into \(y\geq\frac{2}{5}x-2\), \(0\geq - 2\), which is true. Another point: \((5,5)\): \(5\geq\frac{2}{5}(5)-2=2 - 2 = 0\), true. Wait, but maybe the correct line is \(y=\frac{1}{2}x-2\)? No, let's go back to the original graph. Wait, the two points on the line: \((-5,-5)\) and \((0,-2)\). Slope \(m=\frac{-2-(-5)}{0 - (-5)}=\frac{3}{5}\). Equation: \(y=\frac{3}{5}x-2\)? No, when \(x = 0\), \(y=-2\); when \(x = 5\), \(y=\frac{3}{5}(5)-2=3 - 2 = 1\), which doesn't match. I think I made a mistake in identifying the points. Let's look at the graph again. The boundary line passes through \((-5,-5)\) and \((5,0)\) (assuming x - axis horizontal, y - axis vertical). The slope is \(\frac{0 - (-5)}{5-(-5)}=\frac{5}{10}=\frac{1}{2}\). The equation is \(y=\frac{1}{2}x-\frac{5}{2}\)? No, when \(x = 5\), \(y=\frac{1}{2}(5)-\frac{5}{2}=0\), correct. When \(x=-5\), \(y=\frac{1}{2}(-5)-\frac{5}{2}=-5\), correct. Now, the shaded region is above the line. So the inequality is \(y\geq\frac{1}{2}x-\frac{5}{2}\). Multiply both sides by 2 to eliminate fraction: \(2y\geq x - 5\), or \(x-2y\leq5\). Wait, let's check with the point \((0,0)\): \(0 - 0\leq5\), \(0\leq5\), true. Another point \((5,5)\): \(5-10\leq5\), \(-5\leq5\), true. The line is solid, so the inequality includes equality. So the inequality is \(x - 2y\leq5\) or \(y\geq\frac{1}{2}x-\frac{5}{2}\).

Step2: Determine the inequality sign

Since the line is solid, the inequality is non - strict (\(\geq\) or \(\leq\)). The shaded region is above the line \(y=\frac{1}{2}x-\frac{5}{2}\), so we use \(y\geq\frac{1}{2}x-\frac{5}{2}\). We can also write it in standard form. Multiply both sides by 2: \(2y\geq x - 5\), then \(x-2y\leq5\).

Answer:

The inequality represented by the graph is \(y\geq\frac{1}{2}x - \frac{5}{2}\) (or \(x - 2y\leq5\))