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.

Question

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=
x +

Answer 1

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

From the points you provided, we can calculate the slope \( m \) using the following formula:

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

Using the points \( (0, 45) \) and \( (4.5, 0) \):

\[ x_1 = 0, \quad y_1 = 45 \] \[ x_2 = 4.5, \quad y_2 = 0 \]

Calculating the slope:

\[ m = \frac{0 - 45}{4.5 - 0} = \frac{-45}{4.5} = -10 \]

Now, we can use one of the points to find the y-intercept \( b \). We'll use the point \( (0, 45) \):

Using the slope-intercept form:

\[ y = mx + b \]

At \( x = 0 \):

\[ 45 = -10(0) + b \]

This simplifies to:

\[ b = 45 \]

Putting this all together, the equation of the trend line is:

\[ y = -10x + 45 \]

So, in summary:

\( y = -10x + 45 \)