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4.1 — Some of This, Some of That
You already know how to read a graph and find the equation of a line. This lesson uses those familiar skills to describe situations that change their behavior partway through.
1. In class: A Develop Understanding Task
Read one graph as several stages
Imagine a traveler’s distance from a starting point as time passes. Sometimes the traveler moves away, sometimes stops, and sometimes moves back. The graph can show all those stages in one picture.
Read the graph left to right. Its corners are \( (0,0),(5,5),(10,5),(15,15),(20,10) \). In this example, the traveler moves away for 5 hours, stays the same distance away for 5 hours, moves away faster for 5 hours, and then returns toward the start for 5 hours.
| Interval of time | Change shown | Slope: change in distance per hour |
|---|---|---|
| \(0\le x\le5\) | Distance rises from 0 to 5 | \(\frac{5-0}{5-0}=1\) |
| \(5<x\le10\) | Distance remains 5 | \(0\) |
| \(10<x\le15\) | Distance rises from 5 to 15 | \(\frac{15-5}{15-10}=2\) |
| \(15<x\le20\) | Distance falls from 15 to 10 | \(\frac{10-15}{20-15}=-1\) |
A piecewise-defined function assigns one formula to one part of its domain and a different formula to another part. Here the graph has four straight sections, so we need four rules. A horizontal section is also a rule: it has a constant value.
The points at times 5 and 10 have the same height, 5. The traveler therefore stays 5 miles from the starting point for those 5 hours. On the next section the slope is 2, so the traveler moves 2 miles farther away each hour. The last section has slope −1, meaning the distance from the start decreases by 1 mile per hour.
A negative slope does not mean negative speed: it describes a decreasing output.
Write a formula for each stage
For a straight line, \(y=mx+b\), where \(m\) is the slope and \(b\) is the value at \(x=0\). You can also find a line’s formula from a known point: \(y-y_1=m(x-x_1)\). We will use whichever is simpler.
First: slope 1 through (0,0), so \(f(x)=x\).
Second: the line is flat at 5, so \(f(x)=5\).
Third: slope 2 through (10,5): \(f(x)-5=2(x-10)\), so \(f(x)=2x-15\).
Fourth: slope −1 through (15,15): \(f(x)-15=-(x-15)\), so \(f(x)=30-x\).
We combine those rules using a piecewise brace, and write next to each rule the interval in which it applies:
The domain is all inputs from 0 through 20. At an interval boundary such as 5, we assign the point to just one displayed rule. Because these adjoining formulas agree at the endpoints, assigning the boundary to the other adjacent rule would give the same output; do not assign the same input to two conflicting outputs.
At \(x=10\), the second rule gives 5. The next rule, if extended to 10, gives \(2(10)-15=5\). The segments meet with no jump. At \(x=15\), \(2(15)-15=15\) and \(30-15=15\), so those pieces meet as well.
Evaluate and solve using function notation
\(f(12)\) means the output when the input is 12. First decide which interval contains 12: \(10<12\le15\). Only then substitute into the corresponding formula.
\[f(12)=2(12)-15=24-15=9.\]
In our story, after 12 hours the traveler is 9 miles from the starting point. Similarly, 7 belongs to the flat section, so \(f(7)=5\). At the transition, \(f(15)=2(15)-15=15\).
The equation \(f(x)=13\) asks a different question: at which inputs does the output equal 13? A single output can occur at several different inputs.
On the increasing third segment: \(2x-15=13\), so \(x=14\), which lies between 10 and 15.
On the decreasing fourth segment: \(30-x=13\), so \(x=17\), which lies between 15 and 20.
Therefore the complete answer is \(x=14\) or \(x=17\). The first piece never reaches 13, and the second stays at 5.
On the first segment, \(x=3\), which is allowed because \(0\le3\le5\). None of the later sections falls to 3. Thus \(x=3\) is the only answer.
The packet gives the graph without meanings or units for its axes. Its points and slopes are mathematical facts, but the story is your choice. The traveler example above is one possible interpretation, not a story supplied by the packet. Pick consistent axes and units in your own story.
2. Ready — Read values from different types of graphs
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Ready problems 1–4 give graphs made from curved and straight parts. You do not need formulas in order to evaluate them. Start at the given input on the horizontal axis, move vertically to the graph, and read the output from the vertical axis. Check the scale: not every grid square necessarily represents one unit.
If the graph has a gap at a particular input, there may be no function value there. An open circle means a point is excluded; a filled circle means it is included. When a curve meets a line or another curve, trace the actual drawn graph rather than extending a nearby section past its endpoint.
Suppose a function follows the line \(y=x+2\) for \(x<1\) and the horizontal line \(y=4\) for \(x\ge1\). To find \(g(-2)\), use the left piece and calculate \(-2+2=0\). To find \(g(3)\), use the right piece and obtain 4. At \(x=1\), the right-hand piece applies, so \(g(1)=4\).
Ready problem 5 asks you to read a real-world graph: Isaac’s distance from home versus hours since school ended. When the line drops, he is getting closer to home. A horizontal line means he is stationary. A steep line means his distance is changing rapidly.
The horizontal piece from \(t=0.5\) to \(t=2.5\) lasts \(2.5-0.5=2\) hours. Its height is 2 miles from home. The flat section from \(t=2.75\) to \(t=3.25\) is the stop at the library, 1.5 miles from home.
At \(t=3\), he is at the library; using \(d(t)\) for distance from home, we write \(d(3)=1.5\). The final walk changes distance by \(1.5\) miles over \(0.5\) hour, a rate of \(1.5/0.5=3\) miles per hour toward home. Compare the absolute values of the slopes when asked which walking stage is fastest.
Check your work — Ready 1–5 answer key
- a–d: \(0,-4,-6,0\). Assumes one unit per grid square; this graph has no labeled scale.
- a–d: \(-3,5,6,2\). Assumes one unit per grid square; this graph has no labeled scale.
- a–d: \(-1,3,3,5\). Assumes one unit per grid square; this graph has no labeled scale.
- a–d: \(2,1,0,2\). Assumes one unit per grid square; this graph has no labeled scale.
- a. 3.75 hours (home at 6:45 pm). b. 2 hours. c. 1.5 miles. d. At the library. e. \(d(3)=1.5\), where \(d(t)\) is distance from home in miles, \(t\) hours after school. f. From 3.25 to 3.75 hours after school (6:15–6:45 pm), at 3 miles per hour.
3. Set — Build a piecewise rule from a situation
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
To write a model, choose the input variable and units, divide the input domain into stages, calculate each stage’s rule, then state exactly when each rule applies. For distance traveled, each new stage usually starts at the total already accumulated.
Parking garage: a flat fee followed by hourly charges (problem 6)
The garage charges $3 for up to the first two hours. After two hours, the charge increases by $2 per additional hour. Under a proportional charge for fractional additional hours interpretation, we obtain:
Here \(x\) is hours and \(p(x)\) is dollars. The expression \(x-2\) measures additional hours beyond the included first two. At 4.5 hours, the charge is \(3+2(4.5-2)=8\) dollars.
The wording “hourly fee is $2 per hour” does not specify how fractions of an extra hour are charged, and does not state whether entering for zero hours costs $0 or $3. The supplied graph supports a straight sloping second piece, so the lesson uses prorated charging after two hours. If the garage rounds up each started hour, the correct real-life function would instead be a step function. We take \(p(0)=0\) if no parking occurs; the shown piecewise formula addresses positive parking times.
Triathlon: accumulate distance from each event (problem 7)
Lexie swims 1 mile in her first hour, cycles another 12 in the second, and runs another 5 in the third. Total distances at the changeovers are 0, 1, 13, and 18 miles.
From hour 0 to 1, the slope is 1: \(L(t)=t\).
From hour 1 to 2, begin at 1 mile and add 12 miles per elapsed hour: \(L(t)=1+12(t-1)\).
From hour 2 to 3, begin at 13 miles and add 5 miles per elapsed hour: \(L(t)=13+5(t-2)\).
Putting the pieces together:
\[L(t)=\begin{cases}t,&0\le t\le1,\\1+12(t-1),&1<t\le2,\\13+5(t-2),&2<t\le3.\end{cases}\]Check: \(L(2)=1+12(1)=13\) and \(L(3)=13+5(1)=18\). Both match the story.
How to check any model
Inspect the slopes, verify the value at every transition, use meaningful input intervals, and make sure the final total agrees with the story. Different formulas may produce the same value at a joining point without making the pieces interchangeable elsewhere.
Check your work — Set 6–7 answer key
- \(p(x)=\begin{cases}3,&0\le x\le2\\3+2(x-2),&x>2\end{cases}\)
- \(L(t)=\begin{cases}t,&0\le t\le1\\12(t-1)+1,&1<t\le2\\5(t-2)+13,&2<t\le3\end{cases}\)
4. Go — Write lines in point-slope form
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The Go homework, problems 8–13, reviews a separate algebra skill: constructing a line’s equation from a slope and point, or from two points. Point-slope form is
Here \(m\) is the line’s slope and \((x_1,y_1)\) is a known point. The signs inside parentheses come from subtraction; negative coordinates turn into additions.
Find the line through \((4,-3)\) with slope \(m=2\).
Substitute \(x_1=4\), \(y_1=-3\), and \(m=2\):
\[y-(-3)=2(x-4)\]\[\boxed{y+3=2(x-4)}.\]We leave this in point-slope form, as the homework requests.
Find the line through \((-2,5)\) and \((4,-7)\). First compute its slope:
\[m=\frac{-7-5}{4-(-2)}=\frac{-12}{6}=-2.\]Use the first point:
\[y-5=-2(x-(-2))\]\[\boxed{y-5=-2(x+2)}.\]Check with the other point: when \(x=4\), the right side is \(-2(6)=-12\); the left side at \(y=-7\) is also \(-7-5=-12\).
Additional fully worked Go practice
Solution: With \(m=-3\), \(x_1=2\), \(y_1=4\),
\[\boxed{y-4=-3(x-2)}.\]Both the sign and the location of the minus sign matter.
Solution, step 1: \(m=\frac{5-(-1)}{3-(-5)}=\frac{6}{8}=\frac34\).
Step 2: use \((-5,-1)\):
\[y-(-1)=\frac34(x-(-5))\]\[\boxed{y+1=\frac34(x+5)}.\]Step 3: check the second point: \(5+1=6\) and \(\frac34(3+5)=6\), so the formula fits both points.
\[m=\frac{2-(-6)}{-4-0}=\frac8{-4}=-2.\]
Using \((0,-6)\), we get \(\boxed{y+6=-2(x-0)}\), or \(y+6=-2x\).
Check your work — Go 8–13 answer key
- \(y-2=3(x-1)\)
- \(y+2=-(x-1)\)
- \(y+1=2(x-5)\)
- \(y-0=-\frac54(x-0)\)
- \(y-7=-2(x+1)\)
- \(y+9=\frac92(x+1)\)
Additional learning resources
These are optional materials selected specifically for piecewise functions and the line-equation review. Links require an internet connection.
| Resource | How it helps | Scope / What to skip |
|---|---|---|
| Khan Academy — Introduction to piecewise functions | Short introduction to the idea of separate rules on intervals (main task). | Close match. Good starting video. |
| Khan Academy — Evaluate piecewise functions (practice) | Practice choosing the appropriate expression for a given input (main task, Ready). | Close match. Focus on evaluating, not later absolute-value lessons. |
| OpenStax Elementary Algebra 2e — 4.6 Find the Equation of a Line | Worked derivation and examples for point-slope form (Go). | Some extra material. Read “Find an Equation of the Line Given the Slope and a Point” and the two-point examples; skip unrelated slope-intercept practice. |
| Professor Leonard — Intermediate Algebra Lecture 8.3: Study of Piecewise Functions and Basic Translations of Graphs | Optional, longer explanation of piecewise functions. | Significantly broader. Watch the piecewise-functions material; the graph-translations portion goes beyond 4.1. Exact segment timestamps could not be verified. |
Teacher’s review — packet-specific considerations
Set problem 6: The parking rate language leaves billing for partial additional hours unspecified, and does not specify a zero-duration charge. The picture suggests a linear continuation after two hours, but the real-world fee policy could instead round to full hours. For a determinate answer, specify: “Charge $3 for a positive stay of at most 2 hours; charge an additional $2 for each extra hour, prorated continuously.”
Main task, problem 1: Axes are intentionally unlabeled because students must create a story. This is a modeling choice, not a mathematical error; interpretations should be evaluated for internal consistency rather than agreement with one fixed context.
No other serious mathematical error was identified in the assigned material.
Coverage of Section 4.1
| Packet component | Skills and related problems | Where taught |
|---|---|---|
| Main task, Part I, 1–4 | Interpret graph, count stages, build linear formulas and intervals, connect story to graph | In class; examples 1–3; packet clarification |
| Main task, Part II, 5–7 | Evaluate \(f(12),f(7),f(15)\); solve \(f(x)=3\) and \(f(x)=13\) | In class; examples 4–6 |
| Ready, 1–4 | Read function values from curved, linear, and multi-part graphs | Section 2; example 7 |
| Ready, 5a–f | Interpret time and distance, horizontal segments, function notation, steepness and speed | Section 2; example 8 |
| Set, 6–7 | Write and interpret parking-fee and cumulative-distance piecewise functions | Section 3; example 9; clarification |
| Go, 8–13 | Point-slope form from slope and point or two points | Section 4; examples 10–11 and Problems A–C |
Source: Mathematics Vision Project, Secondary Math II, Module 4 (2017), “More Functions, More Features,” Section 4.1 and Ready, Set, Go 4.1 (packet pp. 1–4; PDF pp. 4–7). This independent companion is not publisher material.
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