To determine if the relationship shown by the data is linear, we can observe the changes in y as x increases.
Here is the data provided:
\[ \begin{array}{|c|c|} \hline x & y \ \hline 1 & -4 \ 7 & -7 \ 13 & -10 \ 19 & -13 \ \hline \end{array} \]
First, let's calculate the differences in y as x changes:
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From \(x = 1\) to \(x = 7\): \(y\) changes from \(-4\) to \(-7\) → change in \(y = -7 - (-4) = -3\)
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From \(x = 7\) to \(x = 13\): \(y\) changes from \(-7\) to \(-10\) → change in \(y = -10 - (-7) = -3\)
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From \(x = 13\) to \(x = 19\): \(y\) changes from \(-10\) to \(-13\) → change in \(y = -13 - (-10) = -3\)
Now let's examine the changes in \(x\):
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From \(x = 1\) to \(x = 7\): change in \(x = 7 - 1 = 6\)
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From \(x = 7\) to \(x = 13\): change in \(x = 13 - 7 = 6\)
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From \(x = 13\) to \(x = 19\): change in \(x = 19 - 13 = 6\)
Since the change in \(y\) is consistent (\(-3\)) for a consistent change in \(x\) (\(6\)), we can conclude that the relationship is linear.
Finding the equation of the line
We can use the slope-intercept form of the linear equation \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Step 1: Calculate the slope \(m\)
The slope \(m\) is given by \(m = \frac{\Delta y}{\Delta x} = \frac{-3}{6} = -\frac{1}{2}\).
Step 2: Use a point to find \(b\)
We can use one of the points. Let's use \((1, -4)\):
\[ y = mx + b \implies -4 = -\frac{1}{2}(1) + b \]
Solving for \(b\):
\[ -4 = -\frac{1}{2} + b \implies b = -4 + \frac{1}{2} = -\frac{8}{2} + \frac{1}{2} = -\frac{7}{2} \]
Final Equation:
Thus, the equation of the line is:
\[ y = -\frac{1}{2}x - \frac{7}{2} \]
This confirms that the relationship between \(x\) and \(y\) is indeed linear.
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