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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 point)
y=
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Answer 1

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

From the points given:

1. Point (0, 45)
2. Point (2, 25)
3. Point (3, 15)
4. Point (4.5, 0)

We'll use the two points, (2, 25) and (3, 15), to calculate the slope. The formula for slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Plugging in the values:

\[ m = \frac{15 - 25}{3 - 2} = \frac{-10}{1} = -10 \]

Now we can use one of the points to find the y-intercept \( b \). Using the point (2, 25):

\[ y = mx + b \implies 25 = -10(2) + b \]

\[ 25 = -20 + b \]

\[ b = 25 + 20 = 45 \]

Now we have both \( m \) and \( b \). The equation of the trend line is:

\[ y = -10x + 45 \]

So the equation in slope-intercept form of the trend line is:

\[ y = -10x + 45 \]