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# 4.7: Linear Function Graphs

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Recognize and use function notation.
• Graph a linear function.
• Change slope and intercepts of function graphs.
• Analyze graphs of real-world functions.

## Introduction - Functions

So far we have used the term function to describe many of the equations we have been graphing, but the concept of a function is extremely important in mathematics. Not all equations are functions. In order to be a function, the relationship between two variables, \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, must map each \begin{align*}x-\end{align*}value to exactly one \begin{align*}y-\end{align*}value.

Visually this means the graph of \begin{align*}y\end{align*} versus \begin{align*}x\end{align*} must pass the vertical line test meaning that a vertical line drawn through the graph of the function must never intersect the graph in more than one place.

## Use Function Notation

When we write functions we often use the notation ‘\begin{align*}f(x)=\end{align*}’ in place of ‘\begin{align*}y=\end{align*}’. \begin{align*}f(x)=\end{align*} is read “\begin{align*}f\end{align*} of \begin{align*}x\end{align*}”.

Example 1

Rewrite the following equations so that \begin{align*}y\end{align*} is a function of \begin{align*}x\end{align*} and written \begin{align*}f(x)\end{align*}.

a. \begin{align*}y = 2x + 5\end{align*}

b. \begin{align*}y = -0.2x + 7 \end{align*}

c. \begin{align*}x = 4y- 5 \end{align*}

d. \begin{align*}9x + 3y = 6\end{align*}

Solution

a. Simply replace \begin{align*}y\end{align*} with \begin{align*}f(x)\end{align*}. \begin{align*}f(x)=2x+5\end{align*}

b. \begin{align*}f(x)=-0.2x+7\end{align*}

c. Rearrange to isolate \begin{align*}y\end{align*}.

\begin{align*} x & = 4y - 5 && \text{Add}\ 5\ \text{to both sides}. \\ x + 5 & = 4y && \text{Divide by}\ 4. \\ \frac{x + 5} {4} & = y \\ f(x) & = \frac{x + 5} {4}\end{align*}

d. Rearrange to isolate \begin{align*}y\end{align*}.

\begin{align*} 9x + 3y & = 6 && \text{Subtract}\ 9x\ \text{from both sides}.\\ 3y & = 6 - 9x && \text{Divide by}\ 3.\\ y & = \frac{6 -x} {3} = 2 - 3x \\ f(x) & = 2 - 3x\end{align*}

You can think of a function as a machine made up from a number of separate processes. For example, you can look at the function \begin{align*}3x + 2\end{align*} and break it down to the following instructions.

• Take a number
• Multiply it by 3

We can visualize these processes like this:

In this case, the number we chose was 2. Multiplied by 3 it becomes 6. When we add 2 our output is 8.

Let's try that again. This time we will put -3 through our machine to get 7.

On the bottom of this process tree you can see what happens when we put the letter \begin{align*}n\end{align*} (the variable used to represent any number) through the function. We can write the results of these processes.

• \begin{align*}f(2) = 8 \end{align*}
• \begin{align*}f(-3) = -7\end{align*}
• \begin{align*}f(n) = 3n + 2\end{align*}

Example 2

A function is defined as \begin{align*}f(x)=6x-36\end{align*}. Evaluate the following:

a. \begin{align*}f(2)\end{align*}

b. \begin{align*}f(0)\end{align*}

c. \begin{align*}f(36)\end{align*}

d. \begin{align*}f(z)\end{align*}

e. \begin{align*}f(p)\end{align*}

Solution

a. Substitute \begin{align*}x= 2\end{align*} into the function \begin{align*}f(x) \end{align*} \begin{align*}f(2) = 6\cdot 2 - 36 = 12 - 36 = - 24\end{align*}

b. Substitute \begin{align*}x = 0\end{align*} into the function \begin{align*}f(x)\end{align*} \begin{align*}f(0) = 6\cdot 0 - 36 = 0 - 36 = - 36\end{align*}

c. Substitute \begin{align*}x = 36\end{align*} into the function \begin{align*}f(x)\end{align*} \begin{align*}f(36) = 6\cdot 36 - 36 = 216 - 36 = 180\end{align*}

d. Substitute \begin{align*}x = z\end{align*} into the function \begin{align*}f(x)\end{align*} \begin{align*}f(z) = 6z + 36\end{align*}

e. Substitute \begin{align*}x = p\end{align*} into the function \begin{align*}f(x)\end{align*} \begin{align*}f(p) = 6p + 36\end{align*}

## Graph a Linear Function

You can see that the notation ‘\begin{align*}f(x)=\end{align*}’ and ‘\begin{align*}y=\end{align*}’ are interchangeable. This means that we can use all the concepts we have learned so far to graph functions.

Example 3

Graph the function \begin{align*} f(x) = \frac{3x + 5} {4}\end{align*}

Solution

We can write this function in slope intercept form (\begin{align*}y=mx+b\end{align*} form).

\begin{align*} f(x) = \frac{3} {4} x + \frac{5} {4} = 0.75x + 1.25\end{align*}

So our graph will have a \begin{align*}y-\end{align*}intercept of (0, 1.25) and a slope of 0.75.

• Remember that this slope rises by 3 units for every 4 units we move right.

Example 4

Graph the function \begin{align*} f(x) = \frac{7(5 - x)} {5}\end{align*}

Solution

This time we will solve for the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} intercepts.

To solve for \begin{align*}y-\end{align*}intercept substitute \begin{align*}x = 0.\end{align*}

\begin{align*} f(0) = \frac{7(5 - 0)} {5} = \frac{35} {5} = 7\end{align*}

To solve for \begin{align*}x-\end{align*}intercept substitute use \begin{align*}f(x) = 0\end{align*}.

\begin{align*} 0 & = \frac{7(5 - x)} {5} && \text{Multiply by}\ 5\ \text{and distribute}\ 7.\\ 5.0 & = 35 -7x && \text{Add}\ 7x\ \text{to both sides:}\\ 7x & = 35 \\ x & = 5 \end{align*}

Our graph has intercepts (0, 7) and (5, 0).

## Arithmetic Progressions

You may have noticed that with linear functions, when you increase the \begin{align*}x\end{align*} value by one unit, the \begin{align*}y\end{align*} value increases by a fixed amount. This amount is equal to the slope. For example, if we were to make a table of values for the function \begin{align*}f(x) = 2x + 3\end{align*} we might start at \begin{align*}x =0\end{align*} then add one to \begin{align*}x\end{align*} for each row.

\begin{align*}x\end{align*} \begin{align*}f(x)\end{align*}
0 3
1 5
2 7
3 9
4 11

Look at the values for \begin{align*}f(x)\end{align*}. They go up by two (the slope) each time. When we consider continually adding a fixed value to numbers, we get sequences like \begin{align*}\{ 3, 5, 7, 9, 11 \ldots \}\end{align*}. We call these arithmetic progressions. They are characterized by the fact that each number is greater than (or lesser than) than the preceding number by a fixed amount. This amount is called the common difference. The common difference can be found by taking two consecutive terms in a sequence and subtracting the first from the second.

Example 5

Find the common difference for the following arithmetic progressions:

a. \begin{align*}\{7, 11, 15, 19\ldots \} \end{align*}

b. \begin{align*}\{12, 1, -10, -21\ldots \} \end{align*}

c. \begin{align*}\{7, 12, 17\ldots \}\end{align*}

Solution

a. \begin{align*}11 - 7 = 4\! \\ 15 - 11 = 4\! \\ 19 - 15 = 4.\end{align*}

The common difference is 4.

b. \begin{align*}1- 12 = -11\end{align*}. The common difference is -11.

c. There are not two consecutive terms here, but we know that to get the term after 7, we would add the common difference. Then to get to 12, we would add the common difference again. Twice the common difference is \begin{align*}12 - 7 = 5\end{align*}. So the common difference is 2.5.

Arithmetic sequences and linear functions are very closely related. You just learned that to get to the next term in a arithmetic sequence you add the common difference to last term. We have seen that with linear functions the function increases by the value of the slope every time the \begin{align*}x-\end{align*}value is increased by one. As a result, arithmetic sequences and linear functions look very similar.

The graph to the right shows the arithmetic progression \begin{align*}\{-2, 0, 2, 4, 6\ldots \}\end{align*} with the function \begin{align*}y = 2x - 4\end{align*}. The fundamental difference between the two graphs is that an arithmetic sequence is discrete while a linear function is continuous.

• Discrete means that the sequence has \begin{align*}x\end{align*} values only at distinct points (the \begin{align*}1^{st}\end{align*} term, \begin{align*}2^{nd}\end{align*} term, etc). The domain is not all real numbers (often it is whole numbers).
• Continuous means that the function has values for all possible values of \begin{align*}x\end{align*}, the integers and also all of the numbers in between. The domain is all real numbers.

We can write a formula for an arithmetic progression. We will define the first term as \begin{align*}a_1\end{align*} and \begin{align*}d\end{align*} as the common difference. The sequence becomes the following.

\begin{align*}a_1, a_1 +d, a_1 +2d, a_1 +3d, \ldots, a_1 + n \cdot d\end{align*}

• To find the second term \begin{align*}(a_2)\end{align*} we take the first term \begin{align*}(a_1)\end{align*} and add \begin{align*}d\end{align*}.
• To find the third term \begin{align*}(a_3)\end{align*} we take the first term \begin{align*}(a_1)\end{align*} and add \begin{align*}2d\end{align*}.
• To find the \begin{align*}n\end{align*}th term \begin{align*}(a_n)\end{align*} we take the first term \begin{align*}(a_1)\end{align*} and add \begin{align*}(n-1)d\end{align*}.

## Analyze Graphs of Real-World Functions

Example 6

Use the diagram below to determine the three decades since 1940 in which the infant mortality rate decreased most.

Let's make a table of the infant mortality rate in the years 1940, 1950, 1960, 1970, 1980, 1990, 2000.

Infant Mortality Rates: United States, 1940-2004
Year Mortality rate (per 100,000) change over decade
1940 47 N/A
1950 30 -17
1960 26 -4
1970 20 -6
1980 13 -7
1990 9 -4
2000 7 -2

Solution

The best performing decades were the 1940s (1940-1950) with a drop of 17 deaths per 100000. The 1970s (1970-1980) with a drop of 7 deaths per 100000. The 1960s (1960-1970) with a drop of 6 deaths per 100000.

## Lesson Summary

• In order for an equation to be a function, the relationship between the two variables, \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, must map each \begin{align*}x-\end{align*}value to exactly one \begin{align*}y-\end{align*}value, or \begin{align*}y=f(x)\end{align*}.
• The graph of a function of \begin{align*}y\end{align*} versus \begin{align*}x\end{align*} must pass the vertical line test. Any vertical line will only cross the graph of the function in one place.
• The sequence of \begin{align*}f(x)\end{align*} values for a linear function form an arithmetic progression. Each number is greater than (or less than) the preceding number by a fixed amount, or common difference.

## Review Questions

1. When an object falls under gravity, it gains speed at a constant rate of 9.8 m/s every second. An item dropped from the top of the Eiffel Tower, which is 300 meters tall, takes 7.8 seconds to hit the ground. How fast is it moving on impact?
2. A prepaid phone card comes with $20 worth of calls on it. Calls cost a flat rate of$0.16 per minute. Write the value of the card as a function of minutes per calls. Use a function to determine the number of minutes you can make with the card.
3. For each of the following functions evaluate:
1. \begin{align*} f(x) = -2x + 3\end{align*}
2. \begin{align*} f(x) = 0.7x + 3.2\end{align*}
3. \begin{align*} f(x) = \frac{5 (2 - x)} {11}\end{align*}
1. \begin{align*}f(-3)\end{align*}
2. \begin{align*}f(7)\end{align*}
3. \begin{align*}f(0)\end{align*}
4. \begin{align*}f(z) \end{align*}
4. Determine whether the following could be graphs of functions.
5. The roasting guide for a turkey suggests cooking for 100 minutes plus an additional 8 minutes per pound.
1. Write a function for the roasting time the given the turkey weight in pounds \begin{align*}(x)\end{align*}.
2. Determine the time needed to roast a 10 lb turkey.
3. Determine the time needed to roast a 27 lb turkey.
4. Determine the maximum size turkey you could roast in \begin{align*}4\frac{1}{2}\ \text{hours}\end{align*}.
6. Determine the missing terms in the following arithmetic progressions.
1. \begin{align*}\left \{-11, 17, \underline{\;\;\;\;}, 73\right \}\end{align*}
2. \begin{align*}\left \{2, \underline{\;\;\;\;}, -4\right \}\end{align*}
3. \begin{align*}\left \{13, \underline{\;\;\;\;}, \underline{\;\;\;\;}, \underline{\;\;\;\;}, 0 \right \}\end{align*}

1. 76.44 m/s
2. \begin{align*}f(x) = 2000 - 16x;\end{align*} 125 minutes
1. 9
2. -11
3. 3
4. \begin{align*} f(z) = -2z + 3 \end{align*}
1. 1.1
2. 8.1
3. 3.2
4. \begin{align*}f(z) = 0.7z + 3.2 \end{align*}
1. 2.27
2. -2.27
3. 0.909
4. \begin{align*} f(z) = \frac{10} {11} - \frac{5} {11}z\end{align*}
1. yes
2. no
3. no
4. yes
1. \begin{align*}f(x) = 8x + 100\end{align*}
2. \begin{align*}180 \ min = 3 \ hrs\end{align*}
3. \begin{align*}316 \ min = 5 \ hrs \ 16 \ min\end{align*}
4. 21.25 lbs.
1. 45
2. -1
3. 9.75, 6.5, 3.25

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