An illustration of a scatterplot graph shows x-axis ranging from 0 to 15 in increments of 1 and y-axis ranging from 0 to 75 in increments of 5. Multiple points are plotted and a line points upward to the right with an arrowhead on the top.

Question

The line passes approximately through left parenthesis 0 comma 25 right parenthesis, left parenthesis 1 comma 30 right parenthesis, left parenthesis 2 comma 35 right parenthesis, left parenthesis 7 comma 60 right parenthesis, and left parenthesis 10 comma 75 right parenthesis. 5 points are plotted 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 calculate the slope \( m \) using two of the points given on the line. Let's use the points \( (0, 25) \) and \( (10, 75) \).

1. Calculate the slope \( m \):

The slope \( m \) is calculated using the formula:

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

Using the points \( (0, 25) \) and \( (10, 75) \):

\[ m = \frac{75 - 25}{10 - 0} = \frac{50}{10} = 5 \]

2. Use one point to find the y-intercept \( b \):

Now that we have the slope \( m = 5 \), we can use one of the points to find the y-intercept \( b \). Let's use the point \( (0, 25) \):

Since \( x = 0 \), \( y \) at this point is directly \( b \):

\[ b = 25 \]

3. Write the equation:

Now that we have both \( m \) and \( b \), we can write the equation of the trend line:

\[ y = 5x + 25 \]

Thus, the equation in slope-intercept form is:

\[ y = 5x + 25 \]