### Graphs of Linear Equations

You’re stranded downtown late at night with only $8 in your pocket, and your home is 6 miles away. Two cab companies serve this area; one charges $1.20 per mile with an additional $1 fee, and the other charges $0.90 per mile with an additional $2.50 fee. Which cab will be able to get you home?

**Graph a Linear Equation**

A rule is a way of writing the relationship between the two quantities we are graphing. In mathematics, we tend to use the words **formula** and **equation** to describe the rules we get when we express relationships algebraically. Interpreting and graphing these equations is an important skill that you’ll use frequently in math.

*A taxi costs more the further you travel. Taxis usually charge a fee on top of the per-mile charge to cover hire of the vehicle. In this case, the taxi charges $3 as a set fee and $0.80 per mile traveled. Here is the equation linking the cost in dollars \begin{align*}(y)\end{align*} to hire a taxi and the distance traveled in miles \begin{align*}(x)\end{align*}*.

\begin{align*}y = 0.8x + 3\end{align*}

Graph the equation and use your graph to estimate the cost of a seven-mile taxi ride.

We’ll start by making a table of values. We will take a few values for \begin{align*}x\end{align*} (0, 1, 2, 3, and 4), find the corresponding \begin{align*}y-\end{align*}values, and then plot them. Since the question asks us to find the cost for a seven-mile journey, we need to choose a scale that can accommodate this.

First, here’s our table of values:

\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|

0 | 3 |

1 | 3.8 |

2 | 4.6 |

3 | 5.4 |

4 | 6.2 |

And here’s our graph:

To find the cost of a seven-mile journey, first we find \begin{align*}x = 7\end{align*} on the horizontal axis and draw a line up to our graph. Next, we draw a horizontal line across to the \begin{align*}y-\end{align*}axis and read where it hits. It appears to hit around half way between \begin{align*}y = 8\end{align*} and \begin{align*}y = 9\end{align*}. Let's call it 8.5.

**A seven mile taxi ride would cost approximately $8.50 ($8.60 exactly).**

Here are some things you should notice about this graph and the formula that generated it:

- The graph is a straight line (this means that the equation is
**linear**), although the function is**discrete**and really just consists of a series of points. - The graph crosses the \begin{align*}y-\end{align*}axis at \begin{align*}y = 3\end{align*} (notice that there’s \begin{align*}a + 3\end{align*} in the equation—that’s not a coincidence!). This is the base cost of the taxi.
- Every time we move
**over**by one square we move**up**by 0.8 squares (notice that that’s also the coefficient of \begin{align*}x\end{align*} in the equation). This is the rate of charge of the taxi (cost per mile). - If we move over by three squares, we move up by \begin{align*}3 \times 0.8\end{align*} squares.

#### Real-World Example: Paying off Debt

A small business has a debt of $500,000 incurred from start-up costs. It predicts that it can pay off the debt at a rate of $85,000 per year according to the following equation governing years in business \begin{align*}(x)\end{align*} and debt measured in thousands of dollars \begin{align*}(y)\end{align*}.

\begin{align*}y = -85x + 500\end{align*}

Graph the above equation and use your graph to predict when the debt will be fully paid.

First, we start with our table of values:

\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|

0 | 500 |

1 | 415 |

2 | 330 |

3 | 245 |

4 | 160 |

Then we plot our points and draw the line that goes through them:

Notice the scale we’ve chosen here. There’s no need to include any points above \begin{align*}y= 500\end{align*}, but it’s still wise to allow a little extra.

Next we need to determine how many years it takes the debt to reach zero, or in other words, what \begin{align*}x-\end{align*}value will make the \begin{align*}y-\end{align*}value equal 0. We know it’s greater than four (since at \begin{align*}x = 4\end{align*} the \begin{align*}y-\end{align*}value is still positive), so we need an \begin{align*}x-\end{align*}scale that goes well past \begin{align*}x = 4\end{align*}. Here we’ve chosen to show the \begin{align*}x-\end{align*}values from 0 to 12, though there are many other places we could have chosen to stop.

To read the time that the debt is paid off, we simply read the point where the line hits \begin{align*}y = 0\end{align*} (the \begin{align*}x-\end{align*}axis). It looks as if the line hits pretty close to \begin{align*}x = 6\end{align*}. So **the debt will definitely be paid off in six years.**

To see more simple examples of graphing linear equations by hand, see the Khan Academy video on graphing lines at http://www.youtube.com/watch?v=2UrcUfBizyw. The narrator shows how to graph several linear equations, using a table of values to plot points and then connecting the points with a line.

**Analyze Graphs of Linear Functions**

We often use graphs to represent relationships between two linked quantities. It’s useful to be able to interpret the information that graphs convey. For example, the chart below shows a fluctuating stock price over ten weeks. You can read that the index closed the first week at about $68, and at the end of the third week it was at about $62. You may also see that in the first five weeks it lost about 20% of its value, and that it made about 20% gain between weeks seven and ten. Notice that this relationship is discrete, although the dots are connected to make the graph easier to interpret.

Analyzing graphs is a part of life - whether you are trying to decide to buy stock, figure out if your blog readership is increasing, or predict the temperature from a weather report. Many graphs are very complicated, so for now we’ll start off with some simple linear conversion graphs. Algebra starts with basic relationships and builds to more complicated tasks, like reading the graph above.

#### Determining Unknown Values Given a Graph

Below is a graph for converting marked prices in a downtown store into prices that include sales tax. Use the graph to determine the cost including sales tax for a $6.00 pen in the store.

To find the relevant price with tax, first find the correct pre-tax price on the \begin{align*}x-\end{align*}axis. This is the point \begin{align*}x=6\end{align*}.

Draw the line \begin{align*}x = 6\end{align*} up until it meets the function, then draw a horizontal line to the \begin{align*}y-\end{align*}axis. This line hits at \begin{align*}y \approx 6.75\end{align*} (about three fourths of the way from \begin{align*}y = 6\end{align*} to \begin{align*}y = 7\end{align*}).

**Conversions using a Graph **

The graph for converting temperature from Fahrenheit to Celsius is shown below. Use the graph to convert the following:

a) \begin{align*}70^\circ\end{align*} Fahrenheit to Celsius

To find \begin{align*}70^\circ\end{align*} Fahrenheit, we look along the Fahrenheit-axis (in other words the \begin{align*}x-\end{align*}axis) and draw the line \begin{align*}x = 70\end{align*} up to the function. Then we draw a horizontal line to the Celsius-axis (\begin{align*}y-\end{align*}axis). The horizontal line hits the axis at a little over 20 (21 or 22).

**\begin{align*}70^\circ\end{align*} Fahrenheit is approximately equivalent to \begin{align*}21^\circ\end{align*} Celsius.**

b) \begin{align*}0^\circ\end{align*} Celsius to Fahrenheit

To find \begin{align*}0^\circ\end{align*} Celsius, we look at the Fahrenheit-axis (the line \begin{align*}y = 0\end{align*}). The function hits the \begin{align*}x-\end{align*}axis just right of 30.

**\begin{align*}0^\circ\end{align*} Celsius is equivalent to \begin{align*}32^\circ\end{align*} Fahrenheit**

### Examples

The graph for converting temperature from Fahrenheit to Celsius is shown below. Use the graph to convert the following:

#### Example 1

\begin{align*}0^\circ\end{align*} Fahrenheit to Celsius

To find \begin{align*}0^\circ\end{align*} Fahrenheit, we just look at the \begin{align*}y-\end{align*}axis. (Don't forget that this axis is simply the line \begin{align*}x = 0\end{align*}.) The line hits the \begin{align*}y-\end{align*}axis just below the half way point between −15 and −20.

**\begin{align*}0^\circ\end{align*} Fahrenheit is approximately equivalent to \begin{align*}-18^\circ\end{align*} Celsius.**

#### Example 2

\begin{align*}30^\circ\end{align*} Celsius to Fahrenheit

To find \begin{align*}30^\circ\end{align*} Celsius, we look up the Celsius-axis and draw the line \begin{align*}y = 30\end{align*} along to the function. When this horizontal line hits the function, we draw a line straight down to the Fahrenheit-axis. The line hits the axis at approximately 85.

**\begin{align*}30^\circ\end{align*} Celsius is approximately equivalent to \begin{align*}85^\circ\end{align*} Fahrenheit.**

### Review

For 1-3, make a table of values for the following equations and then graph them.

- \begin{align*}y = 2x + 7\end{align*}
- \begin{align*}y = 0.7x - 4\end{align*}
- \begin{align*}y = 6 - 1.25x\end{align*}

*“Think of a number. Multiply it by 20, divide the answer by 9, and then subtract seven from the result.”*- Make a table of values and plot the function that represents this sentence.
- If you picked 0 as your starting number, what number would you end up with?
- To end up with 12, what number would you have to start out with?

- At the airport, you can change your money from dollars into euros. The service costs $5, and for every additional dollar you get 0.7 euros.
- Make a table for this and plot the function on a graph.
- Use your graph to determine how many euros you would get if you give the office $50.
- To get 35 euros, how many dollars would you have to pay?
- The exchange rate drops so that you can only get 0.5 euros per additional dollar. Now how many dollars do you have to pay for 35 euros?

For 6-9, the graph below shows a conversion chart for converting between weight in kilograms and weight in pounds. Use it to convert the following measurements.

- 4 kilograms into weight in pounds
- 9 kilograms into weight in pounds
- 12 pounds into weight in kilograms
- 17 pounds into weight in kilograms

For 10-12, use the graph from problems 6-9 to answer the following questions.

- An employee at a sporting goods store is packing 3-pound weights into a box that can hold 8 kilograms. How many weights can she place in the box?
- After packing those weights, there is some extra space in the box that she wants to fill with one-pound weights. How many of those can she add?
- After packing those, she realizes she misread the label and the box can actually hold 9 kilograms. How many more one-pound weights can she add?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 4.3.