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 \((0, -2)\) and \((2, 0)\) (we can also choose other points, but these are easy to identify).

Step2: Use the slope formula

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\), where \((x_1,y_1)=(0, -2)\) and \((x_2,y_2)=(2, 0)\).
Substituting the values into the formula: \(m=\frac{0 - (-2)}{2 - 0}=\frac{0 + 2}{2}=\frac{2}{2}=1\). Wait, let's check another pair of points to be sure. Let's take \((0, -2)\) and \((6, 6)\). Then \(m=\frac{6 - (-2)}{6 - 0}=\frac{8}{6}=\frac{4}{3}\)? Wait, no, maybe I made a mistake in the first points. Wait, looking at the graph again, when \(x = 0\), \(y=-2\); when \(x = 1\), \(y=-1\); when \(x = 2\), \(y = 0\); when \(x=3\), \(y = 1\), etc. Wait, let's take two clear points. Let's take \((0, -2)\) and \((3, 1)\). Then \(y_2 - y_1=1-(-2)=3\), \(x_2 - x_1=3 - 0 = 3\), so \(m=\frac{3}{3}=1\)? Wait, no, wait when \(x = 0\), \(y=-2\); when \(x = 1\), \(y=-1\) (so the change in \(y\) is \(1\), change in \(x\) is \(1\)), so slope is \(\frac{1}{1}=1\)? Wait, but when \(x = 6\), \(y = 6\)? Wait, no, the line at \(x = 6\) is at \(y = 6\)? Wait, the y - axis at \(x = 6\), the line is at \(y = 6\)? Wait, no, let's re - examine. The line crosses the y - axis at \((0,-2)\) and when \(x = 2\), \(y = 0\), when \(x = 4\), \(y = 2\), when \(x = 6\), \(y = 4\)? Wait, I think I made a mistake earlier. Let's take two correct points. Let's take \((0,-2)\) and \((3,1)\). Then \(y_2 - y_1=1-(-2)=3\), \(x_2 - x_1=3 - 0 = 3\), so \(m = 1\)? Wait, no, wait the line: from \((0,-2)\) to \((1,-1)\): \(\frac{-1-(-2)}{1 - 0}=\frac{1}{1}=1\). From \((1,-1)\) to \((2,0)\): \(\frac{0 - (-1)}{2 - 1}=\frac{1}{1}=1\). From \((2,0)\) to \((3,1)\): \(\frac{1 - 0}{3 - 2}=\frac{1}{1}=1\). So the slope is \(1\)? Wait, but let's check another way. The slope - intercept form is \(y=mx + b\), where \(b=-2\) (the y - intercept). Let's take a point \((x,y)\) on the line, say \((2,0)\). Substitute into \(y=mx - 2\): \(0 = 2m-2\), then \(2m=2\), so \(m = 1\).

Answer:

\(1\)