Example4.11.5

Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.

\(x\) \(y\)
\(0\) \(-4\)
\(1\) \(-6\)
\(2\) \(-8\)
\(3\) \(-10\)

We consider how the values change from one row to the next. From row to row, the \(x\)-value increases by \(1\text{.}\) Also, the \(y\)-value decreases by \(2\) from row to row.

\(x\) \(y\)
\(0\) \(-4\)
\({}+1\rightarrow\) \(1\) \(-6\) \(\leftarrow{}-2\)
\({}+1\rightarrow\) \(2\) \(-8\) \(\leftarrow{}-2\)
\({}+1\rightarrow\) \(3\) \(-10\) \(\leftarrow{}-2\)

Since row-to-row change is always \(1\) for \(x\) and is always \(-2\) for \(y\text{,}\) the rate of change from one row to another row is always the same: \(-2\) units of \(y\) for every \(1\) unit of \(x\text{.}\)

We know that the output for \(x = 0\) is \(y = -4\text{.}\) And our observation about the constant rate of change tells us that if we increase the input by \(x\) units from \(0\text{,}\) the ouput should decrease by \(\overbrace{(-2)+(-2)+\cdots+(-2)}^{x\text{ times}}\text{,}\) which is \(-2x\text{.}\) So the output would be \(-4-2x\text{.}\)

So the equation is \(y=-2x-4\text{.}\)

in-context