###### Example4.4.5Matthew's Savings

On Dec. 31, Matthew had only \(\$50\) in his savings account. For the the new year, he resolved to deposit \(\$20\) into his savings account each week, without withdrawing any money from the account.

Matthew keeps his resolution, and his account balance increases steadily by \(\$20\) each week. That's a constant rate of change, so his account balance and time have a linear relationship with slope \(20\, \sfrac{\text{dollars}}{\text{wk}}\text{.}\)

We can model the balance, \(y\text{,}\) in Matthew's savings account after \(x\) weeks with an equation. Since Matthew started with \(\$50\) and adds \(\$20\) each week, the account balance \(y\) after \(x\) weeks is

\begin{equation} y = 50 + 20x\tag{4.4.2} \end{equation}where \(y\) is a dollar amount. Notice that the slope, \(20\, \sfrac{\text{dollars}}{\text{wk}}\text{,}\) serves as the multiplier for \(x\) weeks.

We can also consider Matthew's savings using a table.

\(x\text{,}\) weeks since Dec. 31 |
\(y\text{,}\) savings account balance (dollars) |
||

\(0\) | \(50\) | ||

\(x\) increases by \(1\)\(\longrightarrow\) | \(1\) | \(70\) | \(\longleftarrow\)\(y\) increases by \(20\) |

\(x\) increases by \(1\)\(\longrightarrow\) | \(2\) | \(90\) | \(\longleftarrow\)\(y\) increases by \(20\) |

\(x\) increases by \(2\)\(\longrightarrow\) | \(4\) | \(130\) | \(\longleftarrow\)\(y\) increases by \(40\) |

\(x\) increases by \(3\)\(\longrightarrow\) | \(7\) | \(190\) | \(\longleftarrow\)\(y\) increases by \(60\) |

\(x\) increases by \(5\)\(\longrightarrow\) | \(12\) | \(290\) | \(\longleftarrow\)\(y\) increases by \(100\) |

In first few rows of the table, we see that when the number of weeks \(x\) increases by \(1\text{,}\) the balance \(y\) increases by \(20\text{.}\) The row-to-row rate of change is \(\frac{20}{1} = 20\text{,}\) the slope. In any table for a linear relationship, whenever \(x\) increases by \(1\) unit, \(y\) will increase by the slope.

In further rows, notice that as row-to-row change in \(x\) increases, row-to-row change in \(y\) increases proportionally to preserve the constant rate of change. Looking at the change in the last two rows of the table, we see \(x\) increases by \(5\) and \(y\) increases by \(20\text{,}\) which gives rate of change \(\frac{100}{5} = 20\text{,}\) the value of the slope again.

We can “see” the rates of change between consecutive rows of the table on a graph of Matthew's savings by including **slope triangles**.

The large, labeled slope triangle indicates that when \(5\) weeks pass, Matthew saves \(\$100\text{.}\) This is the rate of change between the last two rows of the table, \(\frac{100}{5} = 20 \, \sfrac{\text{dollars}}{\text{wk}}\text{.}\)

The smaller slope triangles indicate, from left to right, the rates of change \(\frac{20}{1}\text{,}\) \(\frac{20}{1}\text{,}\) \(\frac{40}{2}\text{,}\) and \(\frac{60}{3}\) respectively. All of these rates simplify to the slope, \(20 \, \sfrac{\text{dollars}}{\text{wk}}\text{.}\)

Every slope triangle on the graph of Matthew's savings has the same shape (geometrically, they are called similar triangles) since the ratio of vertical change to horizontal change is always \(20 \, \sfrac{\text{dollars}}{\text{wk}}\text{.}\) On any graph of any line, we can draw a slope triangle and compute slope as “rise over run.”

Of course, we could draw a slope triangle on the other side of the line:

This slope triangle works just as well for identifying “rise” and “run,” but it focuses on vertical change before horizontal change. For consistency with mathematical conventions, we will generally draw slope triangles showing horizontal change followed by vertical change, as in Figure 4.4.7.