Question

use the following graph to find x when y = -1. when y = -1, x = box

Explanation:

Step1: Identify the line's equation (optional, but helps).

First, find two points on the line. From the graph, when \( x = 0 \), \( y = 1 \) (y-intercept), and when \( x = 2 \), \( y = -1 \)? Wait, no, let's check again. Wait, when \( y = -1 \), we need to find x. Alternatively, use the graph: draw a horizontal line at \( y = -1 \), find where it intersects the blue line, then find the x-coordinate of that intersection.

Step2: Locate \( y = -1 \) on the y-axis.

Draw a horizontal line (parallel to x-axis) at \( y = -1 \). This line will intersect the blue line. Now, find the x-coordinate of that intersection point.

Looking at the graph, the blue line passes through points. Let's list some points: when \( x = 0 \), \( y = 1 \); when \( x = 1 \), \( y = 0 \); when \( x = 2 \), \( y = -1 \)? Wait, no, let's check the grid. Wait, the grid has x from -2 to 4, y from -6 to 6. Let's see the slope. From (0,1) to (2,-1): slope \( m = \frac{-1 - 1}{2 - 0} = \frac{-2}{2} = -1 \). So equation is \( y = -x + 1 \).

Now, set \( y = -1 \): \( -1 = -x + 1 \). Solve for x: \( -x = -1 - 1 = -2 \), so \( x = 2 \)? Wait, no, wait: \( -1 = -x + 1 \) → \( -x = -2 \) → \( x = 2 \). Wait, but let's check the graph. Wait, when x=2, y=-1? Let's see the graph: the blue line at x=2, y is -1? Wait, the point (2, -1) is on the line? Wait, the graph shows the blue line going through (0,1), (1,0), (2,-1)? Wait, no, maybe I misread. Wait, the grid: each square is 1 unit. Let's check the intersection at y=-1.

Wait, maybe my initial point was wrong. Let's re-express: the line passes through (0,1) and (2,-1)? Wait, no, when x=2, the y-coordinate on the line: looking at the graph, the blue line at x=2 is at y=-1? Wait, the graph's blue line: when x=0, y=1; when x=1, y=0; when x=2, y=-1. So when y=-1, x=2. Wait, but let's confirm with the graph. If we draw y=-1, the horizontal line, it intersects the blue line at x=2. So x=2.

Wait, but let's check again. Wait, maybe I made a mistake. Wait, the line: from (0,1) to (2,-1): yes, that's correct. So when y=-1, x=2.

Answer:

\( \boxed{2} \)