Question

question
graph the line $y = -\frac{2}{3}x + 8$ with appropriate labels and scale. plot the point on the line that represents the solution to the question in part c.

Explanation:

Step1: Identify the y - intercept

The equation of the line is in slope - intercept form \(y = mx + b\), where \(b\) is the y - intercept. For the line \(y=-\frac{2}{3}x + 8\), when \(x = 0\), \(y=8\). So the y - intercept is the point \((0,8)\). We plot this point on the y - axis.

Step2: Identify the slope

The slope \(m=-\frac{2}{3}\). The slope is \(\frac{\text{rise}}{\text{run}}\), which means from the y - intercept \((0,8)\), we move down 2 units (because the numerator of the slope is - 2, negative means down) and then move 3 units to the right (because the denominator of the slope is 3, positive means right). This gives us the next point \((0 + 3,8-2)=(3,6)\). We can also move up 2 units and left 3 units from the y - intercept to get another point \((0 - 3,8 + 2)=(-3,10)\).

Step3: Draw the line

Using the points we found (e.g., \((0,8)\) and \((3,6)\)), we draw a straight line passing through these points. We also need to label the axes (x - axis and y - axis) with appropriate scales. For example, we can mark the x - axis and y - axis with intervals of 1 unit.

Step4: Plot the solution point (if part C is known)

Since part C is not provided, but assuming part C gives an \(x\) (or \(y\)) value, we substitute it into the equation \(y =-\frac{2}{3}x + 8\) to find the corresponding \(y\) (or \(x\)) value and then plot that point on the line we just drew.

(Note: Since the actual graphing is a visual task, the above steps describe how to graph the line. If we were to graph it, we would use the points and the slope to create the line on the given coordinate plane.)

Answer:

To graph \(y =-\frac{2}{3}x+8\):

1. Plot the y - intercept \((0,8)\).
2. Use the slope \(-\frac{2}{3}\) to find another point (e.g., from \((0,8)\), move 3 right and 2 down to get \((3,6)\)).
3. Draw a line through the plotted points.
4. Plot the solution point from part C (after finding its coordinates using the line's equation).