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Trend Lines Practice

Complete this assessment to review what you've learned. It will not count toward your grade.
4 of 54 of 5 Items

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 30 in increments of 2. Multiple points are plotted around a line that points upward to the right with an arrowhead on the bottom.
The line passes approximately through left parenthesis 0 comma 12 right parenthesis, left parenthesis 4 comma 16 right parenthesis, and left parenthesis 6 comma 18 right parenthesis. 12 points are plotted close to the line.

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

(1 point)
y=

1 answer

To write the equation of the trend line in slope-intercept form (\(y = mx + b\)), you first need to determine the slope (\(m\)) and the y-intercept (\(b\)).

From the points you provided, we can choose two of them to calculate the slope. Let's use the points (0, 12) and (4, 16):

  1. Calculate the slope (\(m\)): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{16 - 12}{4 - 0} = \frac{4}{4} = 1 \]

  2. Use the slope and one point to find the y-intercept (\(b\)): We can use the point (0, 12): \[ y = mx + b \implies 12 = (1)(0) + b \implies b = 12 \]

  3. Write the equation in slope-intercept form: Now that we have the slope and intercept, the equation is: \[ y = 1x + 12 \quad \text{or} \quad y = x + 12 \]

So, in slope-intercept form, the equation of the trend line is: \[ y = x + 12 \]