Question

what is the slope of this line? simplify your answer and write it as a proper fraction, improper fraction, or integer.

Explanation:

Step1: Identify two points on the line

From the graph, we can see that the line passes through the points \((0, -1)\) and \((1, 0)\) (we can also use other points, but these are easy to identify).

Step2: Use the slope formula

The slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1, y_1)=(0, -1)\) and \((x_2, y_2)=(1, 0)\). Then \(m=\frac{0 - (-1)}{1 - 0}=\frac{0 + 1}{1}=\frac{1}{1} = 1\)? Wait, no, let's check another pair. Wait, maybe I made a mistake. Let's take two other points. Let's take \((0,-1)\) and \((2,1)\). Then \(y_2 - y_1=1-(-1)=2\), \(x_2 - x_1=2 - 0 = 2\), so \(m=\frac{2}{2}=1\)? Wait, no, wait the line: when \(x = 0\), \(y=-1\); when \(x = 1\), \(y = 0\); when \(x=2\), \(y = 1\); when \(x = 3\), \(y=2\). So the change in \(y\) (rise) is \(1\) when change in \(x\) (run) is \(1\). Wait, but let's check the point \((-8, -7)\) and \((0, -1)\). Then \(y_2 - y_1=-1-(-7)=6\), \(x_2 - x_1=0 - (-8)=8\)? No, that can't be. Wait, no, maybe my first points are wrong. Wait, the line crosses the \(y\)-axis at \((0, -1)\) and the \(x\)-axis at \((1, 0)\)? Wait, no, when \(x = 1\), \(y = 0\)? Let's count the grid. Each grid is 1 unit. So from \((0, -1)\) to \((1, 0)\): up 1, right 1. So slope is \(\frac{1}{1}=1\)? Wait, but let's check another point. From \((0, -1)\) to \((2, 1)\): up 2, right 2, slope \(\frac{2}{2}=1\). From \((-1, -2)\) to \((0, -1)\): up 1, right 1, slope 1. So the slope is 1? Wait, no, wait the line: when \(x = 0\), \(y=-1\); when \(x = 1\), \(y = 0\); so the slope is \(\frac{0 - (-1)}{1 - 0}=\frac{1}{1}=1\). Wait, but let's check the point \((-8, -7)\): when \(x=-8\), \(y=-7\). Then from \((-8, -7)\) to \((0, -1)\): \(y\) changes by \(-1-(-7)=6\), \(x\) changes by \(0 - (-8)=8\), so \(\frac{6}{8}=\frac{3}{4}\)? Wait, that's a contradiction. Wait, no, I must have misread the graph. Wait, the line: let's look again. The line goes through \((0, -1)\) and \((4, 3)\)? Wait, no, the graph: the \(y\)-axis is from -8 to 8, \(x\)-axis from -8 to 8. Let's find two clear points. Let's take \((0, -1)\) and \((4, 3)\). Then \(y_2 - y_1=3 - (-1)=4\), \(x_2 - x_1=4 - 0 = 4\), so slope is \(\frac{4}{4}=1\). Wait, or \((-4, -5)\) and \((0, -1)\): \(y\) change is \(-1-(-5)=4\), \(x\) change is \(0 - (-4)=4\), slope 1. So maybe my initial thought about \((-8, -7)\) was wrong. So the slope is 1? Wait, no, wait the line: when \(x = 0\), \(y=-1\); when \(x = 1\), \(y = 0\); so the slope is \(\frac{0 - (-1)}{1 - 0}=1\). So the slope is 1.

Wait, maybe I made a mistake. Let's use the formula correctly. The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points: \((0, -1)\) and \((2, 1)\). Then \(y_2 - y_1=1 - (-1)=2\), \(x_2 - x_1=2 - 0 = 2\), so \(m=\frac{2}{2}=1\). Another pair: \((1, 0)\) and \((3, 2)\): \(y_2 - y_1=2 - 0 = 2\), \(x_2 - x_1=3 - 1 = 2\), so \(m=\frac{2}{2}=1\). So the slope is 1.

Answer:

\(1\)