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Secondary Math I · Module 3 · Features of Functions

3.6 — Interpreting Functions

Supplement for Mathematics Vision Project (MVP) curriculum · Secondary Math I, Module 3, Section 3.6
Independently created to support students using MVP Math. Not official publisher material and not affiliated with or endorsed by the Mathematics Vision Project.

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A function can be described by a graph, a table, a rule, or a situation. This lesson practices translating between those descriptions: finding inputs and outputs, comparing functions, describing their features, and making a graph that meets given conditions. Use it alongside Section 3.6 of your MVP packet.

1. In class: A Practice Understanding Task

Try the packet problems on your own first, then use these visible answers to check your understanding. The graph and table are in the packet; they are not repeated here.

Problems 1–8 — Read the graph

In your MVP packetSection 3.6 · Main task · p. 27 · Problems 1–8

Answers 1–8

  1. \(f(2)=6\)
  2. \(f(x)=3\) at \(x=0\)
  3. \((-2,0)\)
  4. \([-4,6]\)
  5. \((-2,6]\)
  6. Increasing on \((-4,6)\)
  7. There is no decreasing interval.
  8. \((0,6]\)

Problems 9–14 — Compare and add functions

In your MVP packetSection 3.6 · Main task · p. 28 · Problems 9–14

Answers 9–13

  1. The graphs meet at \(t=-2\) and \(t=3\), at \((-2,-1)\) and \((3,4)\).
  2. \((-2,3)\). Between the intersections, the line \(f\) is above the curve \(g\).
  3. \(f(0)+g(0)=1+(-5)=-4\)
  4. \(f(-1)+g(-1)=0+(-4)=-4\)
  5. \(g(-3)\) is greater: \(f(0)=1\), while \(g(-3)=4\).

14 — Points for the sum graph on \([-1,3]\)

At each input, add the two outputs from the graphs. For example, at \(t=2\), \(f(2)+g(2)=3+(-1)=2\).

\(t\)\(f(t)\)\(g(t)\)\(f(t)+g(t)\)
−10−4−4
01−5−4
12−4−2
23−12
3448

Plot these points and connect them with the upward-curving shape of the sum.

Problems 15–19 — Read the table

In your MVP packetSection 3.6 · Main task · p. 28 · Problems 15–19

Answers 15–19

  1. \(g(-3)=-5\)
  2. \(f(x)=0\) at \(x=1\) and \(x=2\)
  3. \(f\) appears to increase from \(x=2\) to \(x=6\).
  4. \(g(x)>f(x)\) from approximately \(x=-0.6\) to \(x=6\); the table brackets the crossing between \(-1\) and \(0\).
  5. \(f\) changes faster: its average rate has magnitude \(9.5\), compared with \(4\) for \(g\).

Problems 20–25 — Interpret the relations

In your MVP packetSection 3.6 · Main task · p. 29 · Problems 20–25

Answers 20–25

  1. \(h(x)=2^x\), \(f(x)=3x-2\), \(g(x)=8\), and \(y=5x+1\) are functions of \(x\). The relation \(x=4\) is not: it gives many \(y\)-values for the same input.
  2. \(f(2)=4\), \(g(2)=8\), and \(h(2)=4\)
  3. \(g(x)+h(x)=8+2^x\)
  4. \(g(x)<h(x)\) when \(x>3\)
  5. \(f(x)=3x-2\) is increasing for all real \(x\).
  6. \(h(x)=2^x\) has the fastest long-term growth; exponential growth eventually exceeds the linear functions.

Problems 26–29 — Construct graphs

In your MVP packetSection 3.6 · Main task · p. 29 · Problems 26–29

These ask you to create graphs, so many answers are possible. Check your graph against each condition in the packet.

2. Ready — Find a line intersection algebraically

In your MVP packetSection 3.6 · Ready · p. 30 · Problems 1–4

Set two rules equal and find their intersection.

At an intersection, both rules produce the same output. If they are already written as functions of x, set their expressions equal, solve for x, and substitute that x into either rule to find y. Check the other rule too. The answer is the ordered pair \((x,y)\).

Worked example 6 — Solve and check both coordinates

Find the intersection of \(f(x)=2x+1\) and \(g(x)=-x+7\).

\[\begin{aligned}2x+1&=-x+7\\3x+1&=7\\3x&=6\\x&=2.\end{aligned}\]

At \(x=2\), \(f(2)=2(2)+1=5\) and \(g(2)=-2+7=5\). Since the question asks for the intersection point, report both coordinates: \((2,5)\), not just \(x=2\).

Worked example 7 — Clear a fractional coefficient

Find the intersection of \(a(x)=\frac12x-1\) and \(b(x)=-x+5\).

\[\begin{aligned}\tfrac12x-1&=-x+5\\x-2&=-2x+10\\3x&=12\\x&=4.\end{aligned}\]

Substitution gives \(a(4)=2-1=1\) and \(b(4)=-4+5=1\). The intersection is \((4,1)\).

Note: In the second step, we multiply both sides by 2 to clear the fraction and make the arithmetic easier. When clearing fractions, multiply every term on both sides, including constant terms.

Check your work — Ready 1–4 answer key

Try the problems on your own before checking these intersection points.

  1. \((5,-13)\)
  2. \((6,5)\)
  3. \((6,1)\)
  4. \((0,-6)\)

3. Set — Read graphs, distance, and population functions

Problems 5–8 — Read graph values, including gaps

In your MVP packetSection 3.6 · Set · p. 31 · Problems 5–8

Read values from the four graphs.

Try Set 5–8 on your own first. Then open the answer key to check your work.

Check your work — Set 5–8 answer key

5. \(f(x)\)

(a) 1; (b) 7; (c) \(x=0\); (d) \(x=-4\)

6. \(g(x)\)

(a) −1; (b) −4; (c) \(x=-3\); (d) \(x=-1\)

7. \(h(x)\)

(a) −5; (b) approximately −0.5; (c) \(x=4\); (d) \(x=2\)

8. \(d(x)\)

(a) 1; (b) 2; (c) \(x=3\); (d) \(x=-6,\ 0,\ 1,\ 5\)

Problems 9–10 — Interpret distance and population functions

In your MVP packetSection 3.6 · Set · p. 32 · Problems 9–10

Interpret distance and population models.

Try problems 9 and 10 on your own first. They ask you to interpret function notation and use an exponential model. Then read through the solutions to check your work.

Problem 9 — Fran’s walking distance

Original problem

“Fran collected data on the number of feet she could walk each second and wrote the following rule to model her walking rate \(d(t)=4t\).”

  1. What is Fran looking for if she writes \(d(12)=\) ___?
  2. In this situation, what does \(d(t)=100\) tell you?
  3. How can the function rule be used to indicate a time of 16 seconds was walked?
  4. How can the function rule be used to indicate that a distance of 200 feet was walked?

Possible wording clarification: The rule models distance, not walking rate. A small wording change would make that clearer: “Fran collected data on the distance she could walk each second and wrote the following rule to model her distance: \(d(t)=4t\).”

Answers

  1. \(d(12)=4(12)=48\). After 12 seconds, Fran has walked 48 feet.
  2. \(d(t)=100\) asks for the time \(t\) when Fran has walked 100 feet.
  3. Write \(d(16)\) to represent the distance after 16 seconds. The rule gives \(d(16)=4(16)=64\) feet.
  4. Write \(d(t)=200\) to represent a distance of 200 feet. Solving \(4t=200\) gives \(t=50\) seconds.

Problem 10 — Mr. Multbank’s population model

Original problem

“Mr. Multbank has developed a population growth model for the rodents in the field by his house. He believes that starting each spring the population can be modeled based on the number of weeks with the function \(p(t)=8\cdot2^t\).”

  1. Find \(p(t)=128\).
  2. Find \(p(4)\).
  3. Find \(p(10)\).
  4. Find the number of weeks it will take for the population to be over 20,000.
  5. In a year with 16 weeks of summer, how many rodents would he expect by the end of the summer using Mr. Multbank’s model? What are some factors that could change the actual result from your estimate?

Answers

  1. When the population is 128: Solve \(8\cdot2^t=128\). Divide by 8 to get \(2^t=16=2^4\), so \(t=4\) weeks.
  2. After 4 weeks: \(p(4)=8\cdot2^4=8\cdot16=128\) rodents.
  3. After 10 weeks: Since \(2^{10}=1024\), \(p(10)=8\cdot1024=8{,}192\) rodents.
  4. More than 20,000: Start with the model inequality \(8\cdot2^t>20{,}000\). Divide both sides by 8: \(2^t>20{,}000/8=2{,}500\). Since \(2^{11}=2{,}048\) is below 2,500 and \(2^{12}=4{,}096\) is above it, the population first exceeds 20,000 after 12 weeks.
  5. After 16 weeks: Substitute 16 into the original model: \(p(16)=8\cdot2^{16}\). Because \(2^{16}=2^{10}\cdot2^6=1{,}024\cdot64=65{,}536\), \(p(16)=8\cdot65{,}536=524{,}288\) rodents. This is what the model predicts; the actual population could differ because of limited food or space, predators, disease, or other changes in conditions.

4. Go — Describe the whole graph

In your MVP packetSection 3.6 · Go · p. 33 · Problems 11–13

Describe graph features.

For a refresher on domain, range, increasing and decreasing intervals, and maximum and minimum values, see the graph-feature review in Section 3.3. Try problems 11–13 on your own, then open the answer key to check your work.

Check your work — Go 11–13 answer key

11

Domain: \([0,13]\). Range: \([1,11]\). Decreasing on \((0,3)\) and \((10,13)\); increasing on \((3,10)\). Minimum: 1 at x = 3. Maximum: 11 at x = 10. The graph does not cross the x-axis.

12

Domain: \((−∞,30]\). Range: \([−24,0)\). Decreasing throughout its domain. Minimum: −24 at x = 30. No maximum; the graph approaches 0 but does not reach it, so it does not cross the x-axis.

13

Domain and range: all real numbers. Increasing throughout. No maximum or minimum. The line crosses both axes.

Additional learning resources

Resource and direct linkWhat it supportsScope / What to skip
Khan Academy — Systems of equations with substitutionWorked explanations of solving for both coordinates of a system’s solution, the point where its lines intersect.Supports Ready. The packet gives both rules as expressions for y, so substitution becomes setting those expressions equal. Khan’s examples use more general equation forms.
Khan Academy — Systems of equations with substitution practicePractice solving systems algebraically and finding both coordinates.Related practice for Ready; some equations require rearranging first.
Khan Academy — Introduction to increasing, decreasing, positive or negative intervalsSeparates graph height from direction and explains interval descriptions.Close match to Main 1–8 and Go.
The Organic Chemistry Tutor — Finding The Point of Intersection of Two Linear Equations With & Without GraphingFind both intersection coordinates using substitution and check the result by graphing.Supports Ready’s algebraic intersection method. Some equations are rearranged into slope-intercept form first.
Paul’s Online Math Notes — Linear Systems with Two VariablesExplanation and worked examples of substitution for line intersections.Use the substitution method. Skip elimination and the later special cases for this lesson.
Paul’s — Linear Systems with Two Variables practiceAdditional systems with worked solutions.Choose substitution problems; equations may first need rearranging to isolate y.

Teacher’s review of the source packet

No substantial mathematical correction is needed for this section. Main 25 benefits from a brief distinction between a line’s constant rate and an exponential’s growing rate; the nearby note provides that guidance.

Coverage — Section 3.6

Packet locationSkills supportedWhere taught
Main 1–8 · p. 27Values, zeros, domain, sign, and directionIn class
Main 9–14 · p. 28Equality, comparisons, and a sum graphIn class
Main 15–19 · p. 28Table readings, patterns, and average ratesIn class
Main 20–25 · p. 29Functions, rules, sums, growth comparisonsIn class
Main 26–29 · p. 29Construct and check graphs from conditionsIn class
Ready 1–4 · p. 30Algebraic line intersections, fractional coefficientsSection 2
Set 5–8 · p. 31Inputs/outputs, scales, multiple solutions and gapsIn class and Section 3
Set 9–10 · p. 32Interpret notation, distance, doubling and thresholdsSection 3
Go 11–13 · p. 33Domain/range, direction, extrema and end behaviorSection 4: brief review and answer key

Source: MVP, Secondary Mathematics One, Module 3: Features of Functions (May 2016), Section 3.6, printed pp. 27–33, and its teacher notes. The worked examples and added diagrams are original. Mathematical notation needs an internet connection to load its renderer.

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