Question

problem 2: part b here is your graph from the previous screen. write an equation that represents the graph.

Explanation:

Step1: Identify two points

From the graph, we can see two clear points: \((0, 5)\) and \((10, 8)\)? Wait, no, let's check the grid. Wait, the first point is at \(x = 0\), \(y = 5\)? Wait, no, the axes: the vertical axis (y - axis) seems to have 0 at the top? Wait, maybe the axes are reversed. Wait, the horizontal axis (x - axis) has 0, 5, 10, 15 from left to right, and the vertical axis (y - axis) has 0, 5, 10, 15 from top to bottom. So the first point is \((0, 5)\) (x = 0, y = 5, but y is measured from top, so maybe it's \((0, 5)\) in standard coordinates if we flip the y - axis). Wait, another point: when x = 10, y = 8? No, let's look at the grid. Wait, the line goes from (0, 5) (top - left) to (15, 0) (bottom - right)? Wait, no, the first point is at x = 0, y = 5 (if we take the top as y = 0, bottom as y = 15). Wait, maybe the coordinates are (x, y) where x is horizontal (0 to 15) and y is vertical (0 at top, 15 at bottom). So the first point is (0, 5) and another point is (10, 8)? No, wait, the line passes through (0, 5) and (10, 8)? No, let's calculate the slope. Wait, maybe the two points are (0, 5) and (15, 0). Let's check: from (0, 5) to (15, 0), the change in y is \(0 - 5=- 5\), change in x is \(15 - 0 = 15\), so slope \(m=\frac{-5}{15}=-\frac{1}{3}\). Wait, no, maybe (0, 5) and (10, 8) is wrong. Wait, looking at the graph, the first point is at (0, 5) (x = 0, y = 5, with y - axis top - down) and another point at (10, 8)? No, the dot is at (10, 8)? Wait, no, the grid lines: each square is 1 unit. So from (0, 5) (x = 0, y = 5, y - axis top is 0, bottom is 15) to (10, 8)? No, that doesn't make sense. Wait, maybe the correct points are (0, 5) and (15, 0). Let's check the slope: \(m=\frac{0 - 5}{15 - 0}=-\frac{1}{3}\). Wait, no, if we take the standard coordinate system (x - axis right, y - axis up), we need to flip the y - axis. So if the top is y = 0 and bottom is y = 15, then in standard coordinates (y up), the y - coordinate would be \(15 - y_{original}\). So the first point (0, 5) in original coordinates is (0, 15 - 5)=(0, 10) in standard coordinates. The second point: if in original coordinates it's (15, 0), then in standard coordinates it's (15, 15 - 0)=(15, 15)? No, that's not right. Wait, maybe I made a mistake. Let's look at the line: it starts at (0, 5) (x = 0, y = 5, with y increasing downward) and ends at (15, 0) (x = 15, y = 0, y increasing downward). So in standard coordinates (y increasing upward), we can let \(y'=15 - y\). Then the first point is (0, 15 - 5)=(0, 10), and the second point is (15, 15 - 0)=(15, 15)? No, that's not on the line. Wait, maybe the two points are (0, 5) and (10, 8) in the original (y - down) coordinates. Wait, the slope between (0, 5) and (10, 8) is \(\frac{8 - 5}{10 - 0}=\frac{3}{10}\), which is positive, but the line is decreasing, so slope should be negative. Ah, right! The line is decreasing, so as x increases, y (in original, y - down) decreases. So in original coordinates, x from 0 to 15, y from 5 to 0. So two points: (0, 5) and (15, 0). Now, let's find the slope \(m=\frac{0 - 5}{15 - 0}=-\frac{1}{3}\). Now, using the slope - intercept form \(y=mx + b\). In original coordinates (y - down), when x = 0, y = 5, so \(b = 5\). So the equation is \(y=-\frac{1}{3}x + 5\). But if we want to write it in standard form (y - up), we can let \(y'=15 - y\), then \(y = 15 - y'\). Substituting into \(y=-\frac{1}{3}x + 5\), we get \(15 - y'=-\frac{1}{3}x + 5\), so \(y'=\frac{1}{3}x + 10\). But that seems complicated. Wait, maybe the problem expects the equation in the original coordinate system (y - down). So with slope \(m =-\frac{1}{3}\) and y - intercept \(b = 5\), the equation is \(y=-\frac{1}{3}x + 5\). Wait, let's check with x = 10: \(y=-\frac{10}{3}+5=\frac{5}{3}\approx1.67\), but in the graph, when x = 10, the dot is at y = 8 (original y - down)? No, I think I messed up the coordinates. Let's start over. Let's assume the horizontal axis is x (0 to 15) and vertical axis is y (0 at top, 15 at bottom). So the line passes through (0, 5) and (10, 8)? No, the line is decreasing, so when x increases, y decreases. So (0, 5) and (15, 0): when x = 0, y = 5; x = 15, y = 0. Slope \(m=\frac{0 - 5}{15 - 0}=-\frac{1}{3}\). Equation: \(y=-\frac{1}{3}x + 5\). Let's check x = 10: \(y=-\frac{10}{3}+5=\frac{5}{3}\approx1.67\), but in the graph, the dot at x = 10 is at y = 8? No, maybe the points are (0, 5) and (10, 8) is wrong. Wait, maybe the coordinates are (x, y) with x from 0 to 15 (left to right) and y from 0 to 15 (bottom to top), so the top is y = 15, bottom is y = 0. Then the first point is (0, 10) (x = 0, y = 10) and the second point is (15, 0) (x = 15, y = 0). Then slope \(m=\frac{0 - 10}{15 - 0}=-\frac{2}{3}\). Equation: \(y=-\frac{2}{3}x + 10\). Let's check x = 5: \(y=-\frac{10}{3}+10=\frac{20}{3}\approx6.67\), no. Wait, maybe the two points are (0, 5) and (10, 8) is incorrect. Wait, the user's graph: the line starts at (0, 5) (x = 0, y = 5, y - axis top is 0, bottom is 15) and goes to (15, 0) (x = 15, y = 0). So the slope is \(\frac{0 - 5}{15 - 0}=-\frac{1}{3}\), and the y - intercept (when x = 0) is 5. So the equation is \(y =-\frac{1}{3}x+5\).

Step2: Write the equation

Using the slope - intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. We found \(m=-\frac{1}{3}\) and \(b = 5\). So the equation of the line is \(y=-\frac{1}{3}x + 5\). We can also write it in standard form: \(x+3y = 15\) (by multiplying both sides by 3: \(3y=-x + 15\), then \(x + 3y=15\)).

Answer:

\(y =-\frac{1}{3}x + 5\) (or \(x + 3y=15\))