Question

1. $y = -\frac{1}{3}x + 3$

Explanation:

Step1: Identify y-intercept

The equation is in slope - intercept form \(y = mx + b\), where \(b\) is the y - intercept. For \(y=-\frac{1}{3}x + 3\), the y - intercept \(b = 3\). So the line crosses the y - axis at \((0,3)\).

Step2: Find another point using slope

The slope \(m=-\frac{1}{3}=\frac{\text{rise}}{\text{run}}\). From the y - intercept \((0,3)\), we can go down 1 unit (since rise is - 1) and then 3 units to the right (run is 3). So we get the point \((0 + 3,3-1)=(3,2)\). We can also go up 1 unit and 3 units to the left: \((0 - 3,3 + 1)=(-3,4)\).

Step3: Plot the points and draw the line

Plot the points \((0,3)\), \((3,2)\) (or \((-3,4)\)) on the coordinate plane and draw a straight line through them.

To graph \(y =-\frac{1}{3}x + 3\):

1. Mark the y - intercept \((0,3)\) on the y - axis.
2. Use the slope \(-\frac{1}{3}\) to find a second point. For example, from \((0,3)\), move 3 units to the right (along the x - axis) and 1 unit down (along the negative y - axis) to get the point \((3,2)\).
3. Draw a straight line connecting \((0,3)\) and \((3,2)\) (and extend it in both directions).

Answer:

The graph of \(y =-\frac{1}{3}x + 3\) is a straight line with a y - intercept at \((0,3)\) and a slope of \(-\frac{1}{3}\), passing through points like \((0,3)\) and \((3,2)\) (or other points found using the slope). When plotted on the given coordinate grid, it will have a negative slope, crossing the y - axis at 3 and decreasing as \(x\) increases.