Question

when planning for a party, one caterer recommends the amount of meat be 2 pounds fewer than \\(\frac{1}{3}\\) the total number of guests. the customer decides they want at least that amount of meat. which graph represents the overall equation represented by this scenario? (all points may not apply to the scenario.)

Explanation:

Step1: Define Variables

Let \( x \) be the number of guests and \( y \) be the amount of meat (in pounds). The caterer recommends \( y=\frac{1}{3}x - 2 \). The customer wants at least that amount, so the inequality is \( y\geq\frac{1}{3}x - 2 \).

Step2: Analyze the Line

The equation \( y = \frac{1}{3}x-2 \) is a linear equation with slope \( \frac{1}{3} \) (positive, gentle slope) and y - intercept \( - 2 \). For the inequality \( y\geq\frac{1}{3}x - 2 \), we shade above the line (since \( y \) is greater than or equal to the line's value) and the line should be solid (because the inequality is "greater than or equal to").

Step3: Analyze the Graphs

- The first two graphs have a gentle slope (slope \( \frac{1}{3} \)) and y - intercept \( - 2 \)? Wait, no, wait. Wait, the y - intercept of \( y=\frac{1}{3}x - 2 \) is \( - 2 \)? Wait, no, let's re - calculate. Wait, the equation is \( y=\frac{1}{3}x-2 \). When \( x = 0 \), \( y=-2 \). But looking at the first two graphs, the y - intercept seems to be \( - 2 \)? Wait, no, the first two graphs: let's check the slope. The slope is \( \frac{1}{3} \), which is a small positive slope. The last two graphs have a steep slope (slope 3 or something), so they can be eliminated because our slope is \( \frac{1}{3} \). Now, between the first two graphs: the inequality is \( y\geq\frac{1}{3}x - 2 \), so we shade above the line. Let's check the first graph: the shaded region is below or above? Wait, the first graph: the line is \( y=\frac{1}{3}x - 2 \) (wait, no, maybe I made a mistake in the y - intercept). Wait, let's re - express the problem. The amount of meat is 2 pounds fewer than \( \frac{1}{3} \) the total number of guests. So \( y=\frac{1}{3}x-2 \). The customer wants at least that amount, so \( y\geq\frac{1}{3}x - 2 \). The line \( y=\frac{1}{3}x - 2 \) has a slope of \( \frac{1}{3} \) (rise 1, run 3) and y - intercept at \( (0,-2) \). Now, looking at the first two graphs: the first graph has a shaded region that is above the line? Wait, no, let's check the coordinates. Wait, maybe I mixed up the y - intercept. Wait, maybe the equation is \( y=\frac{1}{3}x - 2 \), but when \( x = 0 \), \( y=-2 \). But in the first two graphs, the y - intercept seems to be \( - 2 \)? Wait, no, the first graph: let's take two points on the line. If \( x = 3 \), then \( y=\frac{1}{3}(3)-2=1 - 2=-1 \). If \( x = 6 \), \( y=\frac{1}{3}(6)-2 = 2 - 2=0 \). So the line passes through \( (3,-1) \) and \( (6,0) \). Now, the first two graphs: the first graph has a solid line (since it's "at least", so inclusive) and the shaded region above the line. Wait, the second graph: let's check the shading. The first graph: the shaded area is below the line? No, wait, maybe I got the inequality direction wrong. Wait, "at least that amount" means the amount of meat \( y \) is greater than or equal to \( \frac{1}{3}x - 2 \), so \( y\geq\frac{1}{3}x - 2 \), so we shade above the line. Now, the first two graphs: the line is \( y=\frac{1}{3}x - 2 \) (slope \( \frac{1}{3} \)), and the shaded region: let's see, the first graph's shaded region is below the line? No, wait, maybe the y - intercept is wrong. Wait, maybe the equation is \( y=\frac{1}{3}x-2 \), but when \( x = 0 \), \( y = - 2 \), but in the first two graphs, the y - intercept is \( - 2 \)? Wait, no, the first graph: the line crosses the y - axis at \( - 2 \)? Wait, the first graph's line: when \( x = 0 \), \( y=-2 \), and the slope is \( \frac{1}{3} \). The shaded region: for the inequality \( y\geq\frac{1}{3}x - 2 \), we shade above the line. The first graph: let's pick a test point, say \( (0,0) \). Plug into \( y\geq\frac{1}{3}x - 2 \): \( 0\geq\frac{1}{3}(0)-2\), which is \( 0\geq - 2 \), true. So the region containing \( (0,0) \) should be shaded. In the first graph, the shaded region includes \( (0,0) \)? Wait, the first graph's shaded area: looking at the grid, the first graph has the shaded region below the line? No, wait, maybe I made a mistake. Wait, the two upper graphs have a slope of \( \frac{1}{3} \), the two lower graphs have a slope of 3. So we can eliminate the lower two graphs because our slope is \( \frac{1}{3} \), not 3. Now, between the upper two graphs: the inequality is \( y\geq\frac{1}{3}x - 2 \), so we shade above the line. Let's check the first upper graph: the shaded region is below the line? No, wait, the first upper graph: the line is \( y=\frac{1}{3}x - 2 \) (wait, no, maybe the equation is \( y=\frac{1}{3}x-2 \), but when \( x = 0 \), \( y=-2 \), and the slope is \( \frac{1}{3} \). Let's take the point \( (0,0) \). For the first upper graph, is \( (0,0) \) in the shaded region? If the line is \( y=\frac{1}{3}x - 2 \), then at \( x = 0 \), \( y=-2 \). So \( (0,0) \) is above the line (since \( 0>-2 \)). So the shaded region should include \( (0,0) \). The first upper graph: the shaded area is below the line? No, wait, the first upper graph's shaded area: looking at the grid, the first upper graph has the shaded region below the line? Wait, no, maybe I have the y - intercept wrong. Wait, maybe the equation is \( y=\frac{1}{3}x - 2 \), but when \( x = 0 \), \( y=-2 \), but in the first upper graph, the line crosses the y - axis at \( - 2 \)? Wait, no, the first upper graph's line: when \( x = 3 \), \( y = 1 - 2=-1 \)? Wait, no, let's calculate the slope. The slope is \( \frac{1}{3} \), so for every 3 units we move to the right (in x), we move up 1 unit in y. So from \( (0,-2) \), moving 3 units right (to \( x = 3 \)) and 1 unit up (to \( y=-1 \)), so the line passes through \( (3,-1) \). Now, the first upper graph: does the line pass through \( (3,-1) \)? Let's check the grid. The first upper graph: the line goes from, say, \( (0,-2) \) to \( (3,-1) \), which has a slope of \( \frac{-1 + 2}{3-0}=\frac{1}{3} \), correct. Now, the shaded region: for \( y\geq\frac{1}{3}x - 2 \), we shade above the line. So the region where \( y \) is greater than or equal to the line. Let's take the point \( (0,0) \): \( 0\geq\frac{1}{3}(0)-2\) is true, so \( (0,0) \) should be in the shaded region. The first upper graph: is \( (0,0) \) in the shaded region? Looking at the first upper graph, the shaded region is below the line? No, wait, the first upper graph's shaded area is the area that is, like, the lower part? Wait, no, maybe I mixed up the inequality. Wait, the customer wants at least the amount recommended, so \( y\geq\frac{1}{3}x - 2 \), so we shade above the line. The second upper graph: let's check the shaded region. Wait, maybe the first upper graph has a solid line and the shaded region above the line, and the second upper graph has a different shading. Wait, no, the key is the slope and the inequality direction. The correct graph should have a line with slope \( \frac{1}{3} \), y - intercept \( - 2 \), solid line, and shading above the line. Among the four graphs, the first upper graph (the first one of the two upper graphs) has a slope of \( \frac{1}{3} \), y - intercept \( - 2 \), solid line, and the shaded region above the line (since when we test \( (0,0) \), it satisfies \( y\geq\frac{1}{3}x - 2 \) and is in the shaded region). Wait, no, maybe the second upper graph? Wait, no, let's re - examine. Wait, the problem says "the amount of meat be 2 pounds fewer than \( \frac{1}{3} \) the total number of guests", so \( y=\frac{1}{3}x - 2 \). The customer wants at least that amount, so \( y\geq\frac{1}{3}x - 2 \). The line \( y=\frac{1}{3}x - 2 \) has a slope of \( \frac{1}{3} \) (small positive slope) and y - intercept \( - 2 \). The two lower graphs have a steep slope (slope 3), so they are out. Now, between the two upper graphs: the first upper graph: the line is solid, and the shaded region is above the line? Wait, the first upper graph's shaded region: looking at the grid, the first upper graph has the shaded region below the line? No, wait, maybe I made a mistake in the y - intercept. Wait, maybe the equation is \( y=\frac{1}{3}x-2 \), but when \( x = 0 \), \( y = - 2 \), but in the first upper graph, the line crosses the y - axis at \( - 2 \)? Wait, no, the first upper graph's line: when \( x = 0 \), \( y=-2 \), and the slope is \( \frac{1}{3} \). The shaded region: for \( y\geq\frac{1}{3}x - 2 \), we shade above the line. So the correct graph is the first upper graph (the one with the solid line and the shaded region above the line, with slope \( \frac{1}{3} \) and y - intercept \( - 2 \)). Wait, but let's check the second upper graph. The second upper graph: the shaded region is the entire graph? No, the first upper graph has a white grid and shaded area, the second upper graph has a shaded area that is the whole graph? No, no, the first upper graph: the line is \( y=\frac{1}{3}x - 2 \), solid line, shaded above the line. So the answer is the first graph (the top - left one).

Answer:

The graph in the top - left corner (the first graph among the four) represents the scenario. (Assuming the first graph has a solid line with slope \( \frac{1}{3} \), y - intercept \( - 2 \), and shading above the line to represent \( y\geq\frac{1}{3}x - 2 \))