]> Chapter 02 Notes

2: Functions, Equations, and Graphs

2-1: Relations and Functions

Relations

A relation is a set of coordinate pairs—some matching between two variables (say, x and y). One of the variables must be labeled as input (or domain; typically x), and the other must be labeled as output (or range; typically y).

A relation can be given as a set of ordered pairs, or in tabular format, or in something called a mapping diagram (the domain and range as lists, with arrows connecting elements).

Any equation in two variables automatically defines a relation—though it may be difficult to determine the domain and range.

Any graph in two dimensions automatically defines a relation. The domain will be all x-values that have a point (every vertical line that touches the graph), and the range will be all y-values that have a point (every horizontal line that touches the graph).

To make a graph of a relation—just plot the points!

Functions

Some relations are more important than others…in particular, if the relation assigns a single y-value (output) to every x-value (input), then the relation is a function. Every x is matched with a single y—though a single y might be attached to several different x-values. For function, x is referred to as the independent variable, and y is referred to as the dependent variable—the idea is that you pick an x-value, then plug it into the function in order to produce a y-value. Since we never just pick a y-value, y is dependent on x.

From a list of ordered pairs, check to see if any two points share the same x-value. If the points are a function, then the y-values must also match.

From a mapping diagram, make sure that connected arrows are touching in the range column. If two arrows are touching in the domain column, then that's not a function.

From a graph, you can use the vertical line test—if there is any vertical line that touches the graph in more than one place, then the graph is not a function.

From an equation, you can almost always determine if the equation represents a function by solving the equation for y. If you can do it without resorting to a ± symbol (or absolute value bars), then you've got a function.

Examples

[1.] Determine the domain and range of the relation ${(1, 5), (2, 10), (3, 15), (4, 20)}$ .

The domain is all x-values: ${1, 2, 3, 4}$ and the range is all y-values: ${5, 10, 15, 20}$ .

[2.] Determine if the following relation is a function.

x	1	1	2	3
y	5	8	13	21

This is not a function, since a single x-value (1) is mapped to two different y-values (5 and 8).

[3.] Determine if the following graph represents a function.

An Image

This is a function—every vertical line touches the graph in only one place.

2-2: Linear Equations

Definitions

A linear equation is an equation (relation) that, when graphed, results in a line. If the equation is a function, then we can call it a linear function. Since there is only one kind of linear equation that isn't also a linear function, we'll focus our attentions on linear functions.

To graph a linear function, you can pick a bunch of x-values, run them through the functions to produce y-values, then plot the resulting points…but that would take quite a bit of time. It turns out that there is a simpler way—you can get the graph of a linear function if you just know its y-intercept and its slope.

The y-intercept of any function is that point where the graph touches the y-axis. Note that these coordinates must be of them form $(0, b)$ , since every point on the y-axis has an x-coordinate of zero. For this reason, we sometimes get lazy, and refer to the y-intercept with just the y-coordinate—just the value b from my generalization above.

The slope of a linear function is a measure of the "steepness" of the line. To find the slope, you need to know the coordinates of two points on the line—call these points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . The slope is defined to be $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

Once you know the slope and y-intercept, plot the y-intercept. Now, move up and over—up (or down) the amount in the numerator of the slope, and over (right) the amount in the denominator of the slope.

Forms

There are many ways to write the equation of a line!

Slope-Intercept Form

$y = m x + b$ , where m is the slope of the line, and b is the y-coordinate of the y-intercept.

Point-Slope Form

$y - y_{1} = m (x - x_{1})$ , where m is the slope of the line, and $(x_{1}, y_{1})$ is any point on the line. If the point you use is the y-intercept, then this form will become the slope-intercept form. This form is incredibly useful and incredibly underappreciated!

Standard Form

$A x + B y = C$ , where A and B are not both equal to zero. This is the only form which can be used for any line, whether it is a function or not. There are some books that will write standard form as $A x + B y + C = 0$ …some will call it General Form…I'll not try to trick you on any distinction between general and standard!

If you solve this equation for y, then you'll have slope-intercept form…and you can just read off the slope. Alternately, the slope will be $- \frac{B}{A}$ , as long as $A \neq 0$ .

Properties

Two lines are parallel if they have the same slope.

Two lines are perpendicular if the product of their slopes is -1, or if one line is vertical and the other is horizontal.

Examples

[4.] Find the slope of the line through the points $(3, 4)$ and $(7, 12)$ .

The slope is $\frac{12 - 4}{7 - 7} = \frac{8}{4} = 2$ .

[5.] Write the equation of the line with slope $- \frac{1}{3}$ that passes through the point $(5, -2)$ .

Use point-slope form: $y - (-2) = - \frac{1}{3} (x - 5)$ . We should solve that for y: $y + 2 = - \frac{1}{3} x + \frac{5}{3} ⇒ - \frac{1}{3} x - \frac{1}{3}$ .

[6.] Write the equation of the line that passes through $(-2, -9)$ and is perpendicular to $4 x - 2 y = 7$ .

The slope of the given line is $\frac{4}{2} = 2$ , so a line that is perpendicular to that must have a slope of $- \frac{1}{2}$ . The equation of the desired line is $y - (-9) = - \frac{1}{2} (x - (-2)) ⇒ y + 9 = - \frac{1}{2} (x + 2) ⇒ y = - \frac{1}{2} x - 10$ .

2-3: Direct Variation

Two variables are in direct variation if they obey an equation of the form $y = k x$ , where $k \in ℝ - {0}$ (k is any real number except for zero). k is called the constant of variation.

Notice that this equation is equivalent to $y = k x + 0$ —thus, the constant of variation is the slope of the graph!

Examples

[7.] For the function given below, determine if y varies directly with x.

x	2	4	6
y	6	12	21

No—the ratio $\frac{y}{x}$ equals 3 for the first two points, but there's a different ratio for the third point ( $\frac{21}{6} = \frac{7}{2} = 3.5$ ).

[8.] x and y vary with direct variation—when $x = 3$ , $y = 4.5$ . What is the equation that relates x and y? What will the value of y be when $x = 8$ ?

The constant of variation is $\frac{y}{x} = \frac{4.5}{3} = 1.5$ , so the equation is $y = 1.5 x$ . When $x = 8$ , $y = 1.5 (8) = 12$ .

2-4: Using Linear Models

The purpose of algebra is to strip away the messy details of reality, so that fundamental relationships can be discovered. This is the reason why word problems seem like such an extra burden—they are! However, there comes a point where you've got to take your pristine relationships and turn them back into something meaningful—what good are all these pretty equations if you can't use them for something? This works both ways, actually—what good are all those real-life measurements you've collected if you can't strip away the details and get a nice algebraic description of the underlying relationship?

Thus, we must briefly delve into the realm of statistics—but just barely; just enough to get the calculator to do some hefty calculations for us. Specifically, we'll have it calculate (what some people call) the line of best fit. This is a linear model of the underlying relationship for data that don't exactly fall on a line. Once the calculator produces this line for us, then we can work with it algebraically as we have been.

So—you've got some data, and you've been asked to develop a linear model. Haul out that calculator, and press STAT. Choose 1:Edit. You get columns, labeled L₁, L₂, etc. Put all the x-values in L₁, and put the y-values in L₂. Press STAT again. Press ► to select CALC. Choose 8: LinReg(a+bx) (you could choose 4, but I have my reasons!). Press ENTER.

Ta-da! If you took my advice and choose option 8, then a is the y-intercept of your line, and b is the slope. In class, we'll talk about how many decimal places to use and how to clear data out of the calculator and all that…

Example

[9.] Here are some data collected by a college student about the heights of her friends and the heights of her friends' boyfriends (all measurements in inches).

Women, x	66	64	66	65	70	65
Men, y	72	68	70	68	74	69

Establish a linear model for the relationship between heights. Predict the height of the boyfriend of a girl who is 68 inches tall.

My computer tells me that the y-intercept is at 1.167 and the slope is 1.045. Thus, the linear model is $y = 1.167 + 1.045 x$ .

When the girl is 68 inches tall, the predicted boyfriend height is $y = 1.167 + 1.045 (68) = 1.167 + 71.0909 = 72.25758$ . This model predicts a boyfriend height of 72.25758 inches.

2-5: Absolute Value Functions and Graphs

Our text refers to a function of the form $f (x) = | m x + b | + c$ as an absolute value function. I wouldn't write it that way…but we can work with it.

Remember that $| m x + b | + c$ reduces to $m x + (b + c)$ when $m x + b \geq 0$ …however, without knowing whether m is positive or negative, we can't actually solve this inequality. We do know that x will be on one side of $- \frac{b}{m}$ . Similarly, $| m x + b | + c$ reduces to $- (m x + b) + c = - m x + (c - b)$ when $m x + b < 0$ …again, this can't be solved, but we do know that x will be on one side of $- \frac{b}{m}$ . So the x-value $- \frac{b}{m}$ is an important point on an absolute value function. When this value is plugged into the function, you get $f (- \frac{b}{m}) = | m (- \frac{b}{m}) + b | + c = | - b + b | + c = c$ —so we will always get a point at $(- \frac{b}{m}, c)$ . This point is called the vertex.

Graphing these by hand is easier once you're looking at an actual problem. The graph will look like a V unless there is a negative sign outside of the absolute value bars—then it will be an upside-down V.

Graphing these on a calculator requires that you use the absolute value function—to find it, press MATH, then press ► to get to NUM. The absolute value function is 1:abs(. Don't forget to close the parentheses where the other vertical bar would go!

Examples

[10.] Sketch $y = | 2 x + 6 | - 1$ by hand.

First, use the definition of absolute value to split this into two equations— $y = 2 x + 6 - 1 = 2 x + 5$ and $y = - (2 x + 6) - 1 = -2 x - 6 - 1 = -2 x - 7$ . Now, graph both of those lightly.

An Image

Since there isn't a negative in front, this should be an upwards-V. Darken that portion for the final graph.

An Image

[11.] Graph $y = 3 x - | 3 x + 5 |$ on your calculator.

This is a strange one!

An Image

(ignore those things that look like little bumps…that graph is perfectly flat over there)

2-6: Vertical and Horizontal Translations

A translation is a movement (horizontal or vertical). Some people call it a slide.

Look at the graphs of $y = x$ and $y = x + 1$ —it looks like you could slide one function up (or down) to make the other function.

Look at the graphs of $y = | x |$ and $y = | x - 2 |$ —it looks like you could slide one of them left (or right) to make the other.

In both cases, we have a parent function (the plain one), and a translated function. For now, the only parent functions we'll look at are $y = x$ , $y = - x$ , $y = | x |$ , and $y = - | x |$ .

For $y = x$ and $y = - x$ , the only meaningful translation is vertical, and it comes from a number added or subtracted on the end. An added number indicates that the graph should e moved up, and a negative number indicates that the graph should be moved down.

For $y = | x |$ and $y = - | x |$ , you can have both vertical and horizontal translations (or both at the same time—which the text calls a diagonal translation). A number added or subtracted inside the absolute value bars will move the graph horizontally, and a number added or subtracted outside the bars will move the graph vertically. There is a template for this: $y = | x - h | + k$ . If you write the function in this form, then h is the amount that the graph should be moved horizontally (positive numbers are to the right, and negative numbers are to the left), and k is the amount the graph should be moved vertically (positive numbers are up and negative numbers are down).

Examples

[12.] Describe how the graph of $y = - x - 5$ can be obtained from the graph of $y = - x$ .

Shift the parent graph ( $y = - x$ ) down 5 units.

[13.] Write the equation of the graph shown below.

An Image

This has been shifted three units to the left…so the equation is $y = | x - (-3) | = | x + 3 |$ .

2-7: Two-Variable Inequalities

If an inequality contains one variable, then the answer is one-dimensional, and can be graphed in one dimension (on a line). For inequalities with two variables, the answer will require two dimensions—a grid.

To solve these, first isolate y (solve for y)—ideally with y on the left. The other side of the inequality should be a nice function (a line or an absolute value function)—go ahead and sketch it lightly. If the inequality is strict, then make your sketch a dotted (dashed) line. If the inequality is inclusive, then make your sketch dark.

Now, look at the direction of the inequality symbol. If it points to the y, then the y-values that solve the inequality are less than those on the graph—thus, shade all y-values below the graph. If the inequality points away from y, then the y-values that solve the inequality are larger than those on the graph—thus, shade all the y-values above the graph.

Examples

[14.] Solve $2 x - 3 y \leq 9$ .

First, isolate y: $2 x - 3 y \leq 9 ⇒ - 3 y \leq 9 - 2 x ⇒ y \geq \frac{2}{3} x - 3$ . Now, graph this as a line—solid, since the inequality is inclusive. The solution is all y-values that are greater than the graph, so shade up.

An Image

[15.] Solve $x + y < 2$ .

Isolate: $x + y < 2 ⇒ y < 2 - x$ . I'll graph this with a dotted line. The solution is all y-values below the graph, so I'll shade down.

An Image

[16.] Solve $y > | x - 1 | + 2$ .

This already has the y isolated, so I just need to graph $y = | x - 1 | + 2$ with a dotted line. The solution is y-values greater than the graph, so I shade up.

An Image

[17.] Solve $y + 3 \leq | x + 4 |$ .

Again, isolate first: $y + 3 \leq | x + 4 | ⇒ y \leq | x + 4 | - 3$ . Graph this one with a solid line, and shade down.

An Image

Page last validated 2010-08-15