4.10: Graphs of Linear Functions
What if you were given an equation like \begin{align*}y = \frac{1}{2}x + 3\end{align*}? How could you write it in function notation, evaluate it for a specific function value, and graph it? After completing this Concept, you'll be able to perform tasks like these.
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CK-12 Foundation: 0410S Linear Function Graphs (H264)
Guidance
The highly exclusive Fellowship of the Green Mantle allows in only a limited number of new members a year. In its third year of membership it has 28 members, in its fourth year it has 33, and in its fifth year it has 38. How many members are admitted a year, and how many founding members were there?
Functions
So far we’ve used the term function to describe many of the equations we’ve been graphing, but in mathematics it’s important to remember that not all equations are functions. In order to be a function, a 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 pronounced “\begin{align*}f\end{align*} of \begin{align*}x\end{align*}”.
Example A
Rewrite the following equations so that \begin{align*}y\end{align*} is a function of \begin{align*}x\end{align*} and is 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): f(x) = 2x + 5\end{align*}
b) Again, replace \begin{align*}y\end{align*} with \begin{align*}f(x): f(x) = -0.2x + 7\end{align*}
c) First we need to solve for \begin{align*}y\end{align*}. Starting with \begin{align*}x = 4y - 5\end{align*}, we add 5 to both sides to get \begin{align*}x + 5 = 4y\end{align*}, divide by 4 to get \begin{align*}\frac{x+5}{4}=y\end{align*}, and then replace \begin{align*}y\end{align*} with \begin{align*}f(x): f(x) = \frac{x+5}{4}\end{align*}.
d) Solve for \begin{align*}y:\end{align*} take \begin{align*}9x + 3y = 6\end{align*}, subtract \begin{align*}9x\end{align*} from both sides to get \begin{align*}3y = 6 - 9x\end{align*}, divide by 3 to get \begin{align*}y=\frac{6-9x}{3}=2-3x\end{align*}, and express as a function: \begin{align*}f(x) = 2 - 3x\end{align*}.
Using the functional notation in an equation gives us more information. For instance, the expression \begin{align*}f(x)=mx+b\end{align*} shows clearly that \begin{align*}x\end{align*} is the independent variable because you plug in values of \begin{align*}x\end{align*} into the function and perform a series of operations on the value of \begin{align*}x\end{align*} in order to calculate the values of the dependent variable, \begin{align*}y\end{align*}.
We can also plug in expressions rather than just numbers. For example, if our function is \begin{align*}f(x)=x+2\end{align*}, we can plug in the expression \begin{align*}(x + 5)\end{align*}. We would express this as \begin{align*}f(x+5)=(x+5)+2=x+7\end{align*}.
Example B
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(z)\end{align*}
d) \begin{align*}f(x + 3)\end{align*}
e) \begin{align*}f(2r - 1)\end{align*}
Solution
a) Substitute \begin{align*}x = 2\end{align*} into the function \begin{align*}f(x): \ 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): \ f(0) = 6 \cdot 0 - 36 = 0 - 36 = - 36\end{align*}
c) Substitute \begin{align*}x = z\end{align*} into the function \begin{align*}f(x): \ f(z) = 6z + 36\end{align*}
d) Substitute \begin{align*}x = (x + 3)\end{align*} into the function \begin{align*}f(x): \ f(x + 3) = 6(x + 3) + 36 = 6x + 18 + 36 = 6x + 54\end{align*}
e) Substitute \begin{align*}x = (2r + 1)\end{align*} into the function \begin{align*}f(x): \ f(2r + 1) = 6(2r + 1) + 36 = 12r + 6 + 36 = 12r + 42\end{align*}
Graph a Linear Function
Since the notations “\begin{align*}f(x) =\end{align*}” and “\begin{align*}y =\end{align*}” are interchangeable, we can use all the concepts we have learned so far to graph functions.
Example C
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*}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.
Arithmetic Progressions
You may have noticed that with linear functions, when you increase the \begin{align*}x-\end{align*}value by 1 unit, the \begin{align*}y-\end{align*}value increases by a fixed amount, 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*} and then add 1 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 |
Notice that the values for \begin{align*}f(x)\end{align*} go up by 2 (the slope) each time. When we repeatedly add a fixed value to a starting number, we get a sequence like {3, 5, 7, 9, 11....}. We call this an arithmetic progression, and it is characterized by the fact that each number is bigger (or smaller) than the preceding number by a fixed amount. This amount is called the common difference. We can find the common difference for a given sequence by taking 2 consecutive terms in the sequence and subtracting the first from the second.
Example D
Find the common difference for the following arithmetic progressions:
a) {7, 11, 15, 19, ...}
b) {12, 1, -10, -21, ...}
c) {7, __, 12, __, 17, ...}
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 2 consecutive terms here, but we know that to get the term after 7 we would add the common difference, and then to get to 12 we would add the common difference again. So twice the common difference is \begin{align*}12 - 7 = 5\end{align*}, and so the common difference is 2.5.
Arithmetic sequences and linear functions are very closely related. To get to the next term in a arithmetic sequence, you add the common difference to the last term; similarly, when the \begin{align*}x-\end{align*}value of a linear function increases by one, the \begin{align*}y-\end{align*}value increases by the amount of the slope. So arithmetic sequences are very much like linear functions, with the common difference playing the same role as the slope.
The graph below shows the arithmetic progression {-2, 0, 2, 4, 6...} along with the function \begin{align*}y = 2x - 4\end{align*}. The only major difference between the two graphs is that an arithmetic sequence is discrete while a linear function is continuous.
We can write a formula for an arithmetic progression: if we define the first term as \begin{align*}a_1\end{align*} and \begin{align*}d\end{align*} as the common difference, then the other terms are as follows:
\begin{align*}&a_1 \qquad \quad a_2 \qquad \qquad a_3 \qquad \qquad a_4 \qquad \qquad \ a_5 \qquad \qquad \qquad \qquad a_n\\ &a_1 \qquad a_1+d \qquad a_1+2d \qquad a_1+3d \qquad a_1+4d \quad \ldots \quad a_1+(n-1) \cdot d\end{align*}
The online calculator at http://planetcalc.com/177/ will tell you the \begin{align*}n\end{align*}th term in an arithmetic progression if you tell it the first term, the common difference, and what value to use for \begin{align*}n\end{align*} (in other words, which term in the sequence you want to know). It will also tell you the sum of all the terms up to that point. Finding sums of sequences is something you will learn to do in future math classes.
Watch this video for help with the Examples above.
CK-12 Foundation: Linear Function Graphs
Vocabulary
- 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.
- 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.
- Functions can be expressed in function notation using \begin{align*}f(x) =\end{align*} in place of \begin{align*}y = \end{align*}.
- 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.
Guided Practice
Graph the function \begin{align*}f(x) = \frac{7(5-x)}{5}\end{align*}.
Solution
This time we’ll solve for the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts.
To solve for the \begin{align*}y-\end{align*}intercept, plug in \begin{align*}x = 0: f(0) = \frac{7(5-0)}{5} = \frac{35}{5} = 7\end{align*}, so the \begin{align*}x-\end{align*}intercept is (0, 7).
To solve for the \begin{align*}x-\end{align*}intercept, set \begin{align*}f(x) = 0: 0 = \frac {7(5-x)}{5}\end{align*}, so \begin{align*}0 = 35 -7x\end{align*}, therefore \begin{align*}7x =35\end{align*} and \begin{align*}x=5\end{align*}. The \begin{align*}y-\end{align*}intercept is (5, 0).
We can graph the function from those two points:
Review Questions
- 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?
- 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 left on the card as a function of minutes used so far.
- Use the function to determine how many minutes of calls you can make with the card.
For questions 3-5, evaluate the function for \begin{align*} f(-3)\end{align*}, \begin{align*} f(0)\end{align*}, \begin{align*} f(z)\end{align*}, \begin{align*}f(x + 3)\end{align*}, \begin{align*}f(2n)\end{align*}, \begin{align*}f(3y + 8)\end{align*}, and \begin{align*}f\left ( \frac{q}{2} \right )\end{align*}.
- \begin{align*}f(x) = -2x+3\end{align*}
- \begin{align*}f(x) = 0.7x+3.2\end{align*}
- \begin{align*}f(x) = \frac {5(2-x)} {11}\end{align*}
For questions 6-9, determine whether the graph could be a functions.
- The roasting guide for a turkey suggests cooking for 100 minutes plus an additional 8 minutes per pound.
- Write a function for the roasting time the given the turkey weight in pounds \begin{align*}(x)\end{align*}.
- Determine the time needed to roast a 10 lb turkey.
- Determine the time needed to roast a 27 lb turkey.
- Determine the maximum size turkey you could roast in 4.5 hours.
For questions 11-13, determine the missing terms in the following arithmetic progressions.
- {-11, 17, __, 73}
- {2, __, -4}
- {13, __, __, __, 0}
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Image Attributions
Here you'll learn how to write equations in function form. You'll also learn how to evaluate and graph linear functions. Finally, you'll learn how to find the common difference of arithmetic progressions.