Trend Lines Practice

Question

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Question
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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=

Answer 1

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

From the description:

1. The trend line seems to be going through the points (0, 45) and (4.5, 0). This gives us two points to calculate the slope.

2. The formula for the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

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

• \((x_1, y_1) = (0, 45)\)
• \((x_2, y_2) = (4.5, 0)\)

Calculating the slope:

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

Now that we have the slope \(m = -10\), we can use one of the points to find the y-intercept \(b\). We'll use the point (0, 45):

Substituting \(m\) and the point into the equation of the line:

\[ y = mx + b \quad \text{(using point (0, 45))} \] \[ 45 = -10(0) + b \quad \Rightarrow \quad 45 = b \]

Now we have both \(m\) and \(b\):

• Slope \(m = -10\)
• Y-intercept \(b = 45\)

Finally, we can write the equation of the trend line:

\[ y = -10x + 45 \]

So, the equation of the trend line is:

\[ \boxed{y = -10x + 45} \]