Question

1. y = ______

m = ____ b = ____
(graph of a line on a coordinate plane with two points marked)

Explanation:

Step1: Identify two points on the line

From the graph, let's find two clear points. Let's say the left point is \((-4, 2)\) and the right point is \((3, 0)\) (we can check the grid: moving from left to right, x increases and y decreases).

Step2: Calculate the slope \(m\)

The formula for slope \(m\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Using the points \((x_1,y_1)=(-4,2)\) and \((x_2,y_2)=(3,0)\):
\(m=\frac{0 - 2}{3 - (-4)}=\frac{-2}{7}=-\frac{2}{7}\)? Wait, maybe I picked the wrong points. Let's re - examine. Wait, maybe the y - intercept is at \((0,1)\)? Wait, no, let's look again. Wait, maybe the two points are \((-3, 2)\) and \((3, 0)\). Let's recalculate: \(m=\frac{0 - 2}{3-(-3)}=\frac{-2}{6}=-\frac{1}{3}\). Wait, maybe the grid is such that each square is 1 unit. Let's take the left point as \((-4, 2)\) and the point where it crosses the y - axis: when \(x = 0\), let's see the y - value. From the line, when \(x=0\), \(y = 1\)? Wait, no, maybe the two points are \((-3, 2)\) and \((3, 0)\). Wait, let's use the slope formula correctly. Let's find two points with integer coordinates. Let's say the left point is \((-3, 2)\) and the right point is \((3, 0)\). Then \(m=\frac{0 - 2}{3-(-3)}=\frac{-2}{6}=-\frac{1}{3}\). Then the y - intercept \(b\): when \(x = 0\), from the line, if we plug \(x = 0\) into \(y=mx + b\), using the point \((3,0)\): \(0=-\frac{1}{3}(3)+b\), \(0=- 1 + b\), so \(b = 1\). Wait, maybe the correct points are \((-4, 2)\) and \((2, 0)\). Then \(m=\frac{0 - 2}{2-(-4)}=\frac{-2}{6}=-\frac{1}{3}\), and \(b\): using \(y=mx + b\), with \((2,0)\): \(0=-\frac{1}{3}(2)+b\), \(b=\frac{2}{3}\)? No, this is confusing. Wait, maybe the two points are \((-3, 2)\) and \((3, 0)\). Let's re - do:

Wait, maybe the line passes through \((-3, 2)\) and \((3, 0)\). Then \(m=\frac{0 - 2}{3-(-3)}=\frac{-2}{6}=-\frac{1}{3}\). Then the equation of the line is \(y=mx + b\). We can use the point \((3,0)\): \(0=-\frac{1}{3}(3)+b\), so \(b = 1\). So the equation is \(y=-\frac{1}{3}x + 1\), \(m=-\frac{1}{3}\), \(b = 1\).

Wait, maybe I made a mistake in point selection. Let's take another approach. Let's look at the line: it goes from the second quadrant to the fourth quadrant, decreasing. Let's find two points with \(x\) and \(y\) as integers. Let's say when \(x=-3\), \(y = 2\); when \(x = 0\), \(y = 1\); when \(x = 3\), \(y = 0\). So the change in \(y\) from \(x=-3\) to \(x = 3\) is \(0 - 2=-2\), change in \(x\) is \(3-(-3)=6\), so slope \(m=\frac{-2}{6}=-\frac{1}{3}\). The y - intercept \(b\) is the value of \(y\) when \(x = 0\), which is \(1\). So the equation is \(y=-\frac{1}{3}x + 1\), \(m =-\frac{1}{3}\), \(b = 1\).

Answer:

\(y=-\frac{1}{3}x + 1\), \(m =-\frac{1}{3}\), \(b = 1\)