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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).
Chapter 6 · pp. 111–168
This chapter shows how to turn real situations into mathematical models, starting with linear functions, which change at a constant rate. You will learn to build, read, and graph linear functions, find intercepts, and work with parallel and perpendicular lines. Later parts cover solving modeling problems, fitting lines to data, and absolute value functions.
Mathematical modeling means giving a real situation a numerical structure, for example a function. Once a situation is described with numbers, we can see patterns, predict what will happen, and make decisions. Many real quantities grow in a few common ways, so we study families of functions, and linear functions come first. (p. 111)
Example: a taxi fare. A taxi charges a 2.67 for each mile. Let be the miles driven and the cost in dollars.
Each mile adds the same amount, so after miles the cost is
The is a fixed charge that does not depend on the input. The is a rate of change, and it is multiplied by the input. (pp. 111–112)
Checking units. A good way to test a model is to check that the units agree. Here the rate is dollars per mile. Dollars per mile times miles gives dollars. That matches the (dollars) and the output (dollars). If the units did not match, the equation would be built wrongly. (p. 112)
The same relationship can be shown in four linked ways:
Try to move easily among all four. Sometimes a problem gives you only some of them. (pp. 112–113)
A linear function is a function whose graph is a line. It can always be written as
Here is the initial value (the output when ), and is the constant rate of change, also called the slope. The form matches how we talk: "it starts at and changes at rate ." For example, a collection with 200 songs that gains 15 songs a month is . (p. 113)
The sign of tells you the direction:
For a line through and , the slope is the change in output divided by the change in input:
For and :
On a line, any two points give the same value. (pp. 114–115)
Common misconception. Some students worry about which point to subtract first. It does not matter, as long as the output and input from the same point are in the same position in both subtractions. Swapping both subtractions flips the sign of the top and the bottom, so the fraction stays the same. The mistake is mixing the order, as in , which gives the wrong sign. (p. 115)
Rate of change from real data. Divide the change in the quantity by the change in time. A city grew from 23,400 to 27,800 people between 2002 and 2006. That is a gain of 4,400 people in 4 years, or 1,100 people per year. The population grew, so a positive slope makes sense. Checking that the sign matches the trend is a quick test of reasonableness. (p. 114)
Read the slope as output units per input unit. Read the intercept as the output when the input is 0. For the diver's pressure model
the slope has units PSI per foot. Pressure rises 0.434 PSI for each extra foot of depth. The initial value is the pressure at the surface (0 feet). (p. 115)
From two points that do not include the vertical intercept. Find first. Then substitute one point into and solve for . Given and , we found , so . Using :
So . (pp. 115–116)
When the point is known. If a graph shows and one other point, you can read directly and only need the slope. A line through and has and
so . (p. 116)
A pay model. An insurance salesperson earned 920 for 5 policies. The slope is dollars per policy. Then gives . So
The intercept is the base pay when no policies are sold. The slope is the commission per policy. (p. 117)
From a table. A table is linear if equal steps in input produce equal changes in output. A rat population table rises by 80 every 2 weeks, which is 40 rats per week. With , the model is . If you do not spot the pattern, compute the slope from any two rows. If the output changes vary, such as 5, 30, 105, 230, the function is not linear. (p. 118)
There are three basic methods:
For , you could plot points, or start at and move right 3, down 2. You could also view it as a stretched, reflected, and shifted copy of . (p. 124)
Graphing by plotting points. Choose inputs that work nicely in the formula. Multiples of the fraction's denominator avoid fractions in the arithmetic. In , use multiples of 3:
This gives the points , , and . Two points are enough to draw the line. A third point is a useful check. Connect the points and extend the line in both directions. (p. 124)
Slope is the vertical change divided by the horizontal change between two points on a line. The vertical change is the rise and the horizontal change is the run, so . Any two points on the line give the same ratio. A slope of means that for every 3 units you move right, the line drops 2 units (p. 125).
This gives a quick way to graph . Plot the vertical intercept , then use the rise and run to reach the next point. Finally, extend the line in both directions. For :
A negative slope can be split as instead. For example, can mean up 3 and left 2. Moving left from the intercept finds points with negative (pp. 125–126).
You can also see as a transformation of the identity function . The number stretches or compresses the graph vertically. If is negative, the graph is also reflected over the horizontal axis. The number shifts the graph up or down, and this shift does not change the slope (pp. 126–127).
Example: is compressed vertically by a factor of , reflected over the -axis, and shifted down 3. The intercept is . Rising 1 and running 2 gives , and the same step again gives . Each step changes by 2 and the output by 1, which matches the slope in size (the reflection flips its sign) (pp. 126–127).
To match equations to graphs, compare features in this order: vertical intercepts, then slope signs, then steepness. Take , , and .
The horizontal intercept is where the graph crosses the horizontal axis, and the output is 0 there. So you find it by solving . For , solve to get . The graph crosses at (p. 128).
Horizontal lines. The output never changes, so the rise is 0 and the slope is 0. The equation is . The line has a vertical intercept but no horizontal intercept, unless it is the line . For example, never touches the -axis (p. 129).
Vertical lines. The input never changes, so the run is 0. The slope would need division by zero, so it is undefined. The equation is . Such a line fails the function test, because one input gives many outputs. It has a horizontal intercept but no vertical intercept, unless it is . For example, crosses the -axis at (p. 129).
Two lines are parallel when their rates of change are equal, . Two vertical lines are also parallel. Such lines never cross unless they are the same line. Any line parallel to has slope 3 (p. 130).
Two lines are perpendicular when they meet at a right angle. In that case their slopes multiply to :
This is the negative reciprocal. A slope of 2 pairs with , a slope of pairs with , and a slope of pairs with (pp. 130–131).
Common mistake: only changing the sign, or only flipping the fraction. You must do both. The perpendicular slope to 3 is , not and not . Check by multiplying: the product must be (pp. 130–131).
First choose the correct slope, then use the given point to solve for . Take the reference line and the point .
If the reference line is given by two points, find its slope first. For and :
The perpendicular slope is 6. Through we get , so and (pp. 131–132).
Two lines that are not parallel cross at exactly one point. To find it, set the formulas equal and solve for the input. Then evaluate either function at that input to get the output. For and :
The lines meet at . Parallel lines meet only if they are the same line, and then they meet at every point (pp. 132–133).
The intersection is also where one function switches from larger to smaller. This helps with inequalities. The line has the larger slope, so it is above to the right of the crossing. Therefore when (p. 133).
Most linear modeling problems can be handled with the same eight steps (pp. 139, 146):
Emily has 400 each week. Let be time in weeks and be money left in dollars. The starting value is 3500. The rate is dollars per week, and it is negative because the money is going down. So (pp. 139–140).
The horizontal intercept is where the money hits zero:
So Emily runs out of money after 8.75 weeks (pp. 139–140).
Real-world trends rarely continue forever, so a model is valid only on a limited domain. Negative time makes no sense. Past 8.75 weeks Emily would be in debt. The reasonable domain is , and the range is (p. 140).
The range depends on the domain you choose. If only whole weeks are allowed and she stops after 8 weeks, then . The domain is and the range becomes (p. 140).
Sometimes two choices each have a linear cost. To decide which is better, write one function for each, find where they are equal, and then check which is smaller on each side of that point.
Example: Jamal can rent a truck from U-Haul (16 up front plus 63 cents a mile). Let be the miles driven.
At 100 miles the two costs are the same. U-Haul's cost grows more slowly (0.59 is less than 0.63), so after 100 miles it falls below Budget. U-Haul is the better choice for trips longer than 100 miles. (p. 141)
When the input is a calendar year, the vertical intercept would sit at year 0, which makes the equation awkward. A better choice is the time since a reference year. Then the first data point has input 0, so the vertical intercept is the starting value. (p. 142)
Example: A town had 6,200 people in 2004 and 8,100 in 2009. Let be years since 2004.
A linear model answers two kinds of questions.
Always turn the answer back into the original context (a year, not just ).
Hard problems are easier when you draw a picture and add helper variables that you do not need in the final answer. (pp. 143–144)
Example: Anna walks east at 4 mph and Emanuel walks south at 3 mph, both starting from the same corner. Their radio range is 2 miles. When do they lose contact?
Notice that the distance is also a linear function. (pp. 143–144)
Geometry questions can often be solved by placing a coordinate system and using line equations. (p. 145)
Example: A straight road runs from Westborough to a town 30 miles east and 10 miles north. A second road, perpendicular to the first, goes to Eastborough, which is 20 miles directly east of Westborough. How far from Westborough is the junction?
The steps are: set coordinates, write the line equations, use the perpendicular slope, find the intersection, and measure the distance. (p. 145)
Real data rarely lie exactly on a line. If a scatter plot looks roughly linear, you can draw a line that seems to fit and write its equation. Doing so lets you describe the trend and make predictions. (p. 151)
Example: Cricket chirps in 15 seconds () rise with temperature. The trend is increasing, so the slope should be positive. A hand-drawn line passes through and . Its slope is and its intercept is 30, so . This is only an approximation because the data points do not sit exactly on the line. (pp. 151–152)
The chirp counts range from 18.5 to 44. Predicting the temperature at 30 chirps is interpolation: °F, which is reasonable. Predicting for fewer than 18.5 or more than 44 chirps is extrapolation. (pp. 152–153)
Extrapolation is riskier. At 40°F, the model gives , so chirps. But crickets usually stop chirping below about 50°F, so the line does not apply there. When a model stops describing reality past some point, this is called model breakdown. (p. 153)
A better fit than eyeballing comes from least-squares regression. It finds the line that makes the sum of the squared vertical gaps between the line and the data points as small as possible. Graphing calculators, spreadsheets, and statistics software can compute it. (pp. 153–154)
For the cricket data it gives . This is close to the hand-drawn line but fits better. At 30 chirps it predicts °F, compared with 66°F from the eyeballed line. (p. 154)
The correlation coefficient is a number from to that shows how closely data follow a line.
The cricket data has , a strong increasing linear relationship. (pp. 154–155)
Common mistake: only makes sense for data with a linear pattern. If the scatter plot clearly curves, tells you nothing useful. Look at the plot first, then compute . (p. 154)
Steps: (1) define a simpler input, such as years since the first year, (2) run the regression, (3) check , (4) use the equation to predict. (pp. 155–156)
Example: US gasoline use from 1994 to 2004. Let be years since 1994.
The absolute value function is the last linear-based function in this chapter. It is piecewise: it is built from two linear pieces.
The second piece turns a negative input into a positive output. For example, and . Outputs are never negative. The graph is a V shape with its corner at the origin, and its two arms have slopes and . (p. 159)
The distance between two numbers and on the number line is . It does not matter which number is larger, because the result is always positive. This lets us write "closeness" statements with absolute value. (pp. 159–160)
Read the number being subtracted as the center. Read the right-hand side as the allowed distance from that center.
The corner point is the most useful feature of the graph. For , the corner is at . The shifts come directly from it. (p. 160)
Example. Suppose the corner is at and the arms have slopes and . The graph is shifted right 3 and down 2, and it is stretched by a factor of 2:
If you cannot read the stretch from the slopes, write and substitute a known point. With the point :
If the graph is flipped over the -axis, is negative. (p. 160)
A stretch can be written two ways. (a vertical stretch) and (a horizontal compression) are the same function, because . (p. 160)
The vertical intercept comes from input , so there is exactly one. The number of horizontal intercepts depends on how the V is shifted and reflected. There can be none, one, or two. For example, a V that opens upward with its corner above the -axis has none. (p. 161)
The function is not one-to-one, since two different inputs can give the same output. So inverses of absolute value functions are usually not discussed. (p. 161)
First isolate the absolute value. Then, for with , the quantity inside must be or . Solve both equations: and . (pp. 161–162)
Example. Find the horizontal intercepts of . Set and isolate: .
So the graph has two horizontal intercepts, at and . (pp. 161–162)
No solution. An absolute value can never equal a negative number. Suppose isolating gives . Then the solution set is empty. Always isolate first, then look at the sign on the other side. (p. 162)
Start by solving the related equation. The absolute value function has no breaks, so its outputs can switch from below a value to above it only by passing through that value. The equality points split the number line into intervals. Then decide which intervals work, either from a graph or by testing one value in each interval. (pp. 162–163)
Example. Solve . The equation gives and . Test one value in each interval:
| Interval | Test | Compare with 4 | |
|---|---|---|---|
| 0 | 5 | greater | |
| 6 | 1 | less | |
| 11 | 6 | greater |
Only the middle interval works, so the solution is , or . (p. 163)
Negative coefficient. Find where is negative. We need . Subtracting 3 gives . Multiplying both sides by reverses the inequality, so . Solving gives and . The graph is reflected downward, so it is negative outside these intercepts: or . In interval notation this is . (p. 164)
"Greater than" gives two outer pieces. For , solving gives and . Testing values shows the outputs are at least 3 outside the gap between them. The solution is or , written . A "less than" inequality gives one inner interval instead. (p. 165)
