Trend Lines Practice

Complete this assessment to review what you've learned. It will not count toward your grade.
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Question
Use the image to answer the question.

An illustration of a scatterplot graph shows x-axis ranging from 0 to 10 in increments of 1 and y-axis ranging from 0 to 50 in increments of 10. Multiple points are plotted around a line that points downward to the right with an arrowhead on the bottom.
The line passes approximately through left parenthesis 0 comma 45 right parenthesis, left parenthesis 2 comma 25 right parenthesis, left parenthesis 3 comma 15 right parenthesis, and left parenthesis 4.5 comma 0 right parenthesis. 13 points are plotted close to the line.

Write an equation in slope-intercept form of the trend line.

1 answer

To write the equation of the trend line in slope-intercept form, we need to find the slope and the y-intercept.

1. Find the slope using two points on the line:

Let's use the points (0, 45) and (4.5, 0).

slope = (y2 - y1) / (x2 - x1)
slope = (0 - 45) / (4.5 - 0)
slope = -45 / 4.5
slope = -10

2. Use the slope and one of the points to find the y-intercept:

Let's use the point (0, 45).

y = mx + b (where m is the slope and b is the y-intercept)
45 = -10(0) + b
45 = b

3. Write the equation in slope-intercept form with the slope and y-intercept:

y = -10x + 45

Therefore, the equation of the trend line in slope-intercept form is y = -10x + 45.