Question

the table shows a function. is the function
x | y
7 | 4
8 1/5 | 6
9 2/5 | 7
linear
nonlinear

Explanation:

Step1: Convert mixed numbers to improper fractions

First, convert \(8\frac{1}{5}\) to an improper fraction: \(8\frac{1}{5}=\frac{8\times5 + 1}{5}=\frac{41}{5}\), and \(9\frac{2}{5}=\frac{9\times5+2}{5}=\frac{47}{5}\). The \(x\)-values are \(7=\frac{35}{5}\), \(\frac{41}{5}\), \(\frac{47}{5}\), and the \(y\)-values are \(4\), \(6\), \(7\).

Step2: Calculate the rate of change (slope) between points

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

- Between \((\frac{35}{5},4)\) and \((\frac{41}{5},6)\):
\(m_1=\frac{6 - 4}{\frac{41}{5}-\frac{35}{5}}=\frac{2}{\frac{6}{5}} = 2\times\frac{5}{6}=\frac{5}{3}\)

- Between \((\frac{41}{5},6)\) and \((\frac{47}{5},7)\):
\(m_2=\frac{7 - 6}{\frac{47}{5}-\frac{41}{5}}=\frac{1}{\frac{6}{5}} = 1\times\frac{5}{6}=\frac{5}{6}\)

Wait, that was a mistake. Wait, let's recalculate the differences in \(x\) and \(y\) correctly. Wait, \(8\frac{1}{5}-7 = 1\frac{1}{5}=\frac{6}{5}\), and \(6 - 4 = 2\). So slope \(m_1=\frac{2}{\frac{6}{5}}=\frac{2\times5}{6}=\frac{5}{3}\)? No, wait, \(2\) divided by \(\frac{6}{5}\) is \(2\times\frac{5}{6}=\frac{5}{3}\approx1.666\). Then \(9\frac{2}{5}-8\frac{1}{5}=1\frac{1}{5}=\frac{6}{5}\), and \(7 - 6 = 1\). So slope \(m_2=\frac{1}{\frac{6}{5}}=\frac{5}{6}\approx0.833\). Wait, that can't be. Wait, no, I think I messed up the \(x\)-differences. Wait, \(8\frac{1}{5}-7 = 1\frac{1}{5}=\frac{6}{5}\), correct. \(9\frac{2}{5}-8\frac{1}{5}=1\frac{1}{5}=\frac{6}{5}\), correct. The \(y\)-differences: \(6 - 4 = 2\), \(7 - 6 = 1\). Wait, that would mean the slopes are different, but that's not right. Wait, maybe I made a mistake in the \(y\)-values. Wait, no, the table is \(x:7, 8\frac{1}{5}, 9\frac{2}{5}\); \(y:4,6,7\). Wait, let's check the differences again. Wait, \(8\frac{1}{5}-7 = 1\frac{1}{5}=\frac{6}{5}\), \(y\) changes from \(4\) to \(6\), so \(\Delta y = 2\). Then \(9\frac{2}{5}-8\frac{1}{5}=1\frac{1}{5}=\frac{6}{5}\), \(y\) changes from \(6\) to \(7\), so \(\Delta y = 1\). Wait, that would imply the slope is changing, but that's not possible. Wait, no, maybe I miscalculated the \(x\)-values. Wait, \(7\) to \(8\frac{1}{5}\): \(8\frac{1}{5}-7 = 1\frac{1}{5}=\frac{6}{5}\), correct. \(8\frac{1}{5}\) to \(9\frac{2}{5}\): \(9\frac{2}{5}-8\frac{1}{5}=1\frac{1}{5}=\frac{6}{5}\), correct. So the \(x\)-differences are both \(\frac{6}{5}\), and the \(y\)-differences are \(2\) and \(1\). Wait, that would mean the slope is not constant, but that's a mistake. Wait, no, maybe the problem is that I misread the \(y\)-values. Wait, the table: first row \(x = 7\), \(y = 4\); second row \(x = 8\frac{1}{5}\), \(y = 6\); third row \(x = 9\frac{2}{5}\), \(y = 7\). Wait, let's calculate the slope between the first and second point: \(\frac{6 - 4}{8\frac{1}{5}-7}=\frac{2}{1\frac{1}{5}}=\frac{2}{\frac{6}{5}}=\frac{5}{3}\approx1.666\). Between the second and third point: \(\frac{7 - 6}{9\frac{2}{5}-8\frac{1}{5}}=\frac{1}{1\frac{1}{5}}=\frac{1}{\frac{6}{5}}=\frac{5}{6}\approx0.833\). Wait, that's different. But that can't be. Wait, maybe I made a mistake in the \(y\)-values. Wait, no, the table is as given. Wait, but maybe the problem is that the \(x\)-differences are the same, and the \(y\)-differences should be proportional. Wait, \(8\frac{1}{5}-7 = 1\frac{1}{5}\), \(9\frac{2}{5}-8\frac{1}{5}=1\frac{1}{5}\), so the \(x\)-step is constant (\(\frac{6}{5}\)). The \(y\)-step: from \(4\) to \(6\) is \(2\), from \(6\) to \(7\) is \(1\). So the \(y\)-step is not constant, so the function is nonlinear? But that contradicts. Wait, no, wait, maybe I miscalculated the \(x\)-values. Wait, \(7\) to \(8\frac{1}{5}\): \(8\frac{1}{5}=8.2\), \(7\) to \(8.2\) is \(1.2\) (which is \(\frac{6}{5}\)). \(8.2\) to \(9.4\) (since \(9\frac{2}{5}=9.4\)) is also \(1.2\). The \(y\)-values: \(4\) to \(6\) is \(2\), \(6\) to \(7\) is \(1\). So the rate of change (slope) is \(2/1.2 = 1.666...\) and \(1/1.2 = 0.833...\), which are different. So the function is nonlinear? But wait, maybe I made a mistake. Wait, let's check again. Wait, the problem says "Is the function linear or nonlinear". A linear function has a constant rate of change (slope). So we need to check if the slope between consecutive points is the same.

First pair: \((7,4)\) and \((8\frac{1}{5},6)\). \(x\)-difference: \(8\frac{1}{5}-7 = 1\frac{1}{5}=\frac{6}{5}\). \(y\)-difference: \(6 - 4 = 2\). Slope: \(2\div\frac{6}{5}=2\times\frac{5}{6}=\frac{5}{3}\approx1.6667\).

Second pair: \((8\frac{1}{5},6)\) and \((9\frac{2}{5},7)\). \(x\)-difference: \(9\frac{2}{5}-8\frac{1}{5}=1\frac{1}{5}=\frac{6}{5}\). \(y\)-difference: \(7 - 6 = 1\). Slope: \(1\div\frac{6}{5}=1\times\frac{5}{6}=\frac{5}{6}\approx0.8333\).

Since \(\frac{5}{3} eq\frac{5}{6}\), the slope is not constant, so the function is nonlinear? Wait, but that seems odd. Wait, maybe I misread the \(y\)-values. Wait, the table: first row \(x = 7\), \(y = 4\); second row \(x = 8\frac{1}{5}\), \(y = 6\); third row \(x = 9\frac{2}{5}\), \(y = 7\). Wait, maybe the \(y\)-values are supposed to be \(4\), \(6\), \(8\)? No, the table shows \(7\). Wait, maybe the problem is that the \(x\)-differences are \(\frac{6}{5}\) each, and the \(y\)-differences should be the same if linear. Since they are not, the function is nonlinear.

Wait, but let's check with another approach. Let's assume the function is linear, so \(y = mx + b\). Using the first point \((7,4)\): \(4 = 7m + b\). Using the second point \((8\frac{1}{5},6)\): \(6 = 8\frac{1}{5}m + b\). Subtract the first equation from the second: \(6 - 4=(8\frac{1}{5}-7)m\), so \(2 = 1\frac{1}{5}m\), so \(m = 2\div1\frac{1}{5}=2\div\frac{6}{5}=\frac{10}{6}=\frac{5}{3}\). Then \(b = 4 - 7\times\frac{5}{3}=4-\frac{35}{3}=\frac{12 - 35}{3}=-\frac{23}{3}\). Now check the third point \((9\frac{2}{5},7)\): \(y=\frac{5}{3}x-\frac{23}{3}\). Plug in \(x = 9\frac{2}{5}=\frac{47}{5}\): \(y=\frac{5}{3}\times\frac{47}{5}-\frac{23}{3}=\frac{47}{3}-\frac{23}{3}=\frac{24}{3}=8\). But the third point has \(y = 7\), which is not equal to \(8\). So the third point does not lie on the line, so the function is nonlinear.

Answer:

nonlinear