A cluster of
Escherichia coli
bacteria, magnified 10,000× by a low-temperature electron microscope. Under good conditions, an E. coli colony doubles roughly every 20 minutes — one cell becomes a billion in under ten hours. Image: Eric Erbe / Christopher Pooley, USDA Agricultural Research Service —
Wikimedia Commons, public domain (U.S. Government work).
A biologist measures a bacteria colony on Monday and again on Wednesday.
Is two measurements enough to predict the population a year from now — exactly?
— or —
What else does the math need to know?
Standards: HSF-LE.A.2, HSF-LE.B.5, HSF-IF.C.8.b. Lesson 6 builds directly on Lesson 5: students just learned that exponential functions have equal factors over equal intervals, even fractional ones. Today they put that property to work — given two points whose inputs do not differ by 1, find the per-unit growth factor and write the equation.
Prep Checklist
☐ No additional materials — everything is on these slides.
☐ Activity 1 uses the Information Gap routine. Pre-arrange pairs and decide who gets the problem card vs. data card. Have the printed cards (slide 5) ready, or project them and assign roles verbally.
☐ Activity 2 (Bacteria Growth Expressions) is optional — skip if pacing is tight. The Lesson Synthesis covers the same skill.
Warm-Up5 min
Do Now
The five expressions below are all equivalent. Discuss with a partner why — using properties of exponents.
\(64^{1/3}\)
\(\sqrt[3]{64}\)
\(\bigl(64^{1/6}\bigr)^{2}\)
\(\bigl(8^{2}\bigr)^{1/3}\)
\(2^{2}\)
Complete on your own. Then confirm your answers with a partner.
Choose one expression, and explain in writing why it equals 4.
Write another exponential equation that also equals 4.
5 min total for warm-up (slides 2–3). Students work for ~3 min, then pair-share. The aim is fluency with non-unit fractional exponents — a skill they will need within seconds of starting Activity 1.
Listen for students who show that each expression cubed is 64 (the cleanest unifying argument), and for those who use the power-of-a-power rule: \(64^{1/3} = 64^{2/6} = (64^{1/6})^{2}\) and \((8^{2})^{1/3} = 8^{2/3} = (8^{1/3})^{2} = 2^{2}\). Either path works.
For Problem 2, expect responses such as \(\sqrt[6]{64^{2}}\), \(\sqrt{16}\), or \(2^{6/3}\). Anything that simplifies to 4 is fine.
Warm-Up · Synthesis5 min
All Equivalent? — Discussion
Each expression is a different name for the same number. Surface two unifying arguments: the cube test, and the power-of-a-power rule.
“Why is \(64^{1/3} = 4\)?”\(64^{1/3}\) means the number whose cube is 64. Since \(4^{3} = 64\), we have \(64^{1/3} = 4\). The notation \(\sqrt[3]{64}\) says the same thing.
“How is \(\bigl(64^{1/6}\bigr)^{2}\) the same as \(64^{1/3}\)?”By the power-of-a-power rule \((b^{m})^{n} = b^{mn}\). So \((64^{1/6})^{2} = 64^{2/6} = 64^{1/3} = 4\). Two sixth-roots compound to one third-root.
“How does \(\bigl(8^{2}\bigr)^{1/3} = 2^{2}\)?”\((8^{2})^{1/3} = 8^{2/3} = (8^{1/3})^{2} = 2^{2} = 4\). Or: \(8^{2} = 64\), and \(64^{1/3} = 4\). Same number, different paths.
“What single test confirms all five at once?”Cube each expression. Each cube equals 64, so each expression must equal \(64^{1/3} = 4\). The cube is the common “fingerprint.”
click to advance discussion ▶
Try Saying
The expression ___ equals 4 because ___.
Try Saying
Two expressions are/aren't equivalent because ___.
Use the discussion to surface that “raising to the 1/3 power” and “taking the cube root” mean the same thing. Today's Activity 1 will require students to take roots of given numbers (e.g. \(b^{2} = 5.0625\) means \(b = 5.0625^{1/2}\), and \(b^{1/2} = 0.9\) means \(b = 0.9^{2}\)). The notation rehearsal here is the foundation.
Activity 1 · Launch3 min
Info Gap: Two Points
You and a partner are each given a card. One of you has a problem.The other has data. The person with the problem must ask for the information they need — without seeing the data card.
★ Problem Card holder
Read your card silently. Decide what information you need.
Ask your partner: “Can you tell me ___?”
Explain why you need it: “I need ___ because ___.”
Keep asking until you can solve the problem.
Solve independently. Then share both cards.
▲ Data Card holder
Read your card silently. Wait for your partner to ask.
Before answering their question, ask them back: “Why do you need to know ___?”
Listen. Give only what is on your card.
Do not solve the problem for them.
After they solve, share both cards and discuss.
For an exponential function \(f(x) = a \cdot b^{x}\), what two pieces of information are enough to write the equation?
Launch (~3 min). Display the protocol cards. Confirm everyone understands the two roles. The closing question seeds the activity: students should land on the initial value (\(a\)) and the per-unit growth factor (\(b\)) — or, equivalently, the \(y\)-intercept and one other point.
Pairing. Arrange pairs. Give one student the problem card; give the other the data card. After Round 1, have students switch roles for Round 2.
Printable cards:unit5_lesson6_info_gap_cards.html — 4 pairs per sheet, 2 sheets total (Round 1 + Round 2). Print, then cut along the dashed lines.
Anchor question (only if students stall): “If I tell you the function passes through \((0, 200)\), what else do I need to tell you to fix the equation?”
Reference · How ToActivity 1 · Work Time15 min
Reference · How To — Equation from two points
Step
What to do
Example
1
Identify a — the starting value, the output when \(x = 0\).
Suppose the graph passes through \((0, 200)\) and \((2, 1012.5)\). Then \(a = 200\).
2
Use the second point to write \(a \cdot b^{n} = y_{2}\), where n is the distance between inputs.
Take the nth root: \(b = \bigl(b^{n}\bigr)^{1/n}\).
\(b = 5.0625^{1/2} = \sqrt{5.0625} = 2.25\).
5
Write the equation \(f(x) = a \cdot b^{x}\).
\(f(x) = 200 \cdot 2.25^{x}\).
★
If \(a\) is not given but the input difference is n, you can still find \(b\): \(b^{n} = \dfrac{y_{2}}{y_{1}}\), so \(b = \!\left(\dfrac{y_{2}}{y_{1}}\right)^{1/n}\).
From \((1, 450)\) and \((3, 2278.125)\): \(b^{2} = \dfrac{2278.125}{450} = 5.0625\), so \(b = 2.25\).
Activity 1 · Information Gap
Round 1 — Problem Card
\(f\) is an exponential function. Find an equation defining \(f(x)\). Show your reasoning.
Round 1 — Data Card
The graph of \(f\) passes through \((0, 200)\) and \((2, 1012.5)\).
Round 2 — Problem Card
\(g\) is an exponential function. Find an equation defining \(g(x)\). Show your reasoning.
Round 2 — Data Card
The \(y\)-intercept of the graph of \(g\) is \(100\), and the graph passes through \(\left(\tfrac{1}{2}, 90\right)\).
15 min · Activity 1 Work Time. Leave this slide on screen during work time. Students should not see the data card unless they have it — project the cards on opposite sides of the room, or use printed handouts.
Round 1 target: \(f(x) = 200 \cdot 2.25^{x}\). Reasoning: \(a = 200\) (the \(y\)-intercept). \(200 \cdot b^{2} = 1012.5\), so \(b^{2} = 5.0625\), so \(b = \sqrt{5.0625} = 2.25\).
The Amplify CL block grades both rounds. The exit input collects the equation only — partners discuss reasoning verbally.
Watch for
Round 2: students often plug the half-step factor 0.9 directly into the equation as \(g(x) = 100 \cdot 0.9^{x}\). That is the half-step factor, not the per-unit factor. Squaring 0.9 to get 0.81 is the move.
Amplify · 2 inputs
Order per round: Text → Math Response (named answer1, answer2) → Note (CL in </>).
Format examples: Example: 200(2.25)^x · Example: 100(0.81)^x
Round 1: Find an equation for f(x). The graph passes through (0, 200) and (2, 1012.5).
content:
when answer1.submitted and answer1.latex = "200(2.25)^{x}"
"Correct! a = 200 (the y-intercept), and b^2 = 1012.5/200 = 5.0625, so b = 2.25."
when answer1.submitted and answer1.latex = "200\cdot(2.25)^{x}"
"Correct!"
when answer1.submitted and answer1.latex = "200 \cdot (2.25)^{x}"
"Correct!"
when answer1.submitted and answer1.latex = "200\cdot2.25^{x}"
"Correct!"
when answer1.submitted and answer1.latex = "200(2.25)^x"
"Correct!"
when answer1.submitted and answer1.latex = "200(5.0625)^{x/2}"
"Equivalent! 5.0625 is the 2-step factor. The per-unit factor is its square root, 2.25 — both equations describe the same function."
when answer1.submitted and answer1.latex = "200(5.0625)^{x}"
"Almost! 5.0625 is the factor for an x-increase of 2, not 1. Take the square root to get the per-unit factor: √5.0625 = 2.25."
when answer1.submitted and answer1.latex = "200(1012.5)^{x}"
"Hmm... 1012.5 is the y-value when x = 2, not the per-unit factor. Use 200 · b^2 = 1012.5 to solve for b."
when answer1.submitted and answer1.latex = "1012.5(2.25)^{x}"
"I see what happened — 1012.5 is the y-value at x = 2, not the y-intercept. The y-intercept (a) is the y-value at x = 0, which is 200."
when answer1.submitted and answer1.latex = "200(2.5)^{x}"
"You're close — check 200 · 2.5^2 = 200 · 6.25 = 1250, not 1012.5. The factor b should satisfy 200 · b^2 = 1012.5. Try b = √5.0625."
when answer1.submitted and not isBlank(answer1.latex)
"Almost — start with a = 200. Then 200 · b^2 = 1012.5 gives b^2 = 5.0625. What is b?"
otherwise ""
Round 2: Find an equation for g(x). y-intercept is 100; graph passes through (1/2, 90).
content:
when answer2.submitted and answer2.latex = "100(0.81)^{x}"
"Correct! a = 100, and 100 · b^(1/2) = 90 gives b^(1/2) = 0.9, so b = 0.9^2 = 0.81."
when answer2.submitted and answer2.latex = "100\cdot(0.81)^{x}"
"Correct!"
when answer2.submitted and answer2.latex = "100\cdot0.81^{x}"
"Correct!"
when answer2.submitted and answer2.latex = "100(0.81)^x"
"Correct!"
when answer2.submitted and answer2.latex = "100(0.9)^{2x}"
"Equivalent! 0.9 is the half-step factor; (0.9)^(2x) = (0.9^2)^x = 0.81^x — same function written two ways."
when answer2.submitted and answer2.latex = "100(0.9)^{x}"
"Not quite — 0.9 is the half-step factor (the factor when x increases by ½), not the per-unit factor. The per-unit factor is 0.9^2 = 0.81."
when answer2.submitted and answer2.latex = "100(90)^{x}"
"Hmm — 90 is the y-value at x = ½, not a growth factor. Use 100 · b^(1/2) = 90 to solve for b."
when answer2.submitted and answer2.latex = "90(0.81)^{x}"
"I see what happened — 90 is the y-value at x = ½, not the y-intercept. The y-intercept (a) is the y-value at x = 0, which is 100."
when answer2.submitted and answer2.latex = "100(0.45)^{x}"
"You're close — check 100 · 0.45^(1/2) = 100 · 0.6708 ≈ 67.1, not 90. The half-step factor is 90/100 = 0.9, so the per-unit factor is 0.9^2 = 0.81."
when answer2.submitted and not isBlank(answer2.latex)
"Almost — start with a = 100. Then 100 · b^(1/2) = 90 gives b^(1/2) = 0.9. What is b?"
otherwise ""
Activity 1 · Synthesis5 min
Two Points — Discussion
Different students likely asked for different information. Surface the strategies, then connect them to a single algebraic move: solve \(a \cdot b^{n} = y_{2}\) for \(b\).
“What two pieces of information did you ask for? Why?”Some asked for the \(y\)-intercept (gives \(a\) directly) and one other point (gives \(b^{n} = y/a\)). Others asked for two non-zero points and used \(b^{n} = y_{2}/y_{1}\). Both work — you need two points, plus the distance between their inputs.
“Why is the distance between inputs (\(n\)) so important?”It tells you what power you took. If \(b^{n} = 5.0625\) and \(n = 2\), then \(b = \sqrt{5.0625} = 2.25\). If you had thought \(n = 1\), you would have written \(b = 5.0625\) — a wrong factor.
“What was the move that surprised you?”For Round 2, the half-step factor \(0.9\) is not the per-unit factor. Two half-steps compound to one whole step, so the per-unit factor is \(0.9^{2} = 0.81\). The pattern: a factor over \(\tfrac{1}{n}\)-units, raised to the \(n\)th power, gives the per-unit factor.
click to advance discussion ▶
Try Saying
To write \(f(x) = a \cdot b^{x}\), I needed ___ and ___ because ___.
Try Saying
The per-unit factor is/is not the same as the factor over the given interval because ___.
Sequence the discussion in this order. The conceptual peak is question 3: students must recognize when the given factor needs to be raised to a power. Round 1: \(b^{2} = 5.0625\), so \(b = 5.0625^{1/2}\). Round 2: \(b^{1/2} = 0.9\), so \(b = 0.9^{2}\). In both cases the rule is the same: \(b = \!\left(\dfrac{y_{2}}{y_{1}}\right)^{1/n}\), where \(n\) is the input distance.
Math As A Second Language
Per-unit growth factor from two points
Math · We Say · Meaning
★ Math (given)
\(b = \!\left(\dfrac{y_{2}}{y_{1}}\right)^{1/n}\)
▲ We Say
“b equals the quantity y-sub-2 over y-sub-1, raised to the one over n”or“the per-unit factor is the n-th root of the ratio of the two outputs”
● Meaning
When two outputs of an exponential function are n input-units apart, the per-unit growth factorb is the n-th root of the ratio \(\dfrac{y_{2}}{y_{1}}\). This formula works for any two points — with or without a known y-intercept.
Standalone MASL slide. The general form \(f(x) = a \cdot b^{x}\) was introduced in earlier lessons; this slide names the formula students just used in Activity 1 to extract \(b\) from two points whose inputs differ by something other than 1. Connect this back to Lesson 5's identity: \(\dfrac{f(x_{1} + n)}{f(x_{1})} = b^{n}\). Solving for b gives the formula above.
Activity 2 · LaunchOptional3 min
Bacteria Growth Expressions
A bacteria population starts at 1,000 and grows exponentially, doubling every 10 hours. A biology textbook gives two different expressions for the population after t hours.
Expression A
\(1000 \cdot 2^{t/10}\)
Expression B
\(1000 \cdot \!\left(2^{1/10}\right)^{t}\)
Which expression looks more familiar? Which one looks easier to read at a glance?
Launch (~3 min). Display both expressions. Ask: which is easier to read? Most students will pick A (the doubling structure is visible). The activity reveals that B has a different reader-friendliness: it shows the per-hour growth factor explicitly. Both describe the same population.
Activity 2 is optional. The Lesson Synthesis (slide 12) reaches the same skill on a graph-driven problem. If pacing is tight, jump to slide 12.
Activity 2 · Work TimeOptional7 min
Population at \(t = 0\): 1,000.
Doubles every 10 hours.
Explain why both expressions represent the bacteria population after t hours.
By what factor does the bacteria population grow each hour? Show or explain how you know.
Try Writing
Both expressions describe the same population because ___. The hourly growth factor is ___ because ___.
7 min · Activity 2 Work Time. Students draft alone (~3 min), then revise with a partner (~3 min) using Stronger and Clearer Each Time on Problem 1. Listening prompts: “What do you mean when you say ___?” and “Can you describe that another way?”
Problem 1: Both equal 1000 at \(t = 0\) (since \(2^{0} = 1\)). Both equal 2000 at \(t = 10\) (since \(2^{10/10} = 2\) and \(\bigl(2^{1/10}\bigr)^{10} = 2^{1} = 2\)). The exponent rule \(2^{t/10} = (2^{1/10})^{t}\) makes them algebraically identical.
Problem 2: Hourly factor = \(2^{1/10} \approx 1.0718\). Reasoning: Expression B shows it directly — whenever t increases by 1, the expression is multiplied by \(2^{1/10}\). Expression A reaches the same conclusion via the exponent rule.
Surface students who say "divide 2 by 10" or "add 0.1" — both miss the multiplicative structure.
Amplify · 2 inputs
Order: Problem 1: Text → Text Response named answer3. Problem 2: Text → Math Response named answer4 → Note (CL in </>).
Format example: Example: 1.0718
Problem 1: Why both expressions describe the same population. [Text Response only]
content:
when answer3.submitted and not isBlank(answer3.content)
"Thank you — share your reasoning during the synthesis discussion."
when answer3.submitted and isBlank(answer3.content)
"Please write a short explanation before submitting."
otherwise ""
Problem 2: Hourly growth factor.
content:
when answer4.submitted and answer4.numericValue > 1.0717 and answer4.numericValue < 1.0719
"Correct! The hourly factor is 2^(1/10) ≈ 1.0718. Expression B shows it directly: each unit increase in t multiplies by 2^(1/10)."
when answer4.submitted and answer4.numericValue > 0.199 and answer4.numericValue < 0.201
"Almost — 0.2 looks like 2 ÷ 10 (linear thinking). The hourly factor compounds: 10 of them multiplied together must equal 2. Try 2^(1/10)."
when answer4.submitted and answer4.numericValue > 1.099 and answer4.numericValue < 1.101
"Hmm — 1.1 (10% per hour) is close, but check: 1.1^10 ≈ 2.59, not 2. The factor that gives 2 after ten compoundings is 2^(1/10) ≈ 1.0718."
when answer4.submitted and answer4.numericValue > 1.999 and answer4.numericValue < 2.001
"Not quite — 2 is the ten-hour factor (the doubling factor), not the hourly factor. The hourly factor compounds 10 times to give 2: it is 2^(1/10) ≈ 1.0718."
when answer4.submitted and answer4.numericValue > 0.0719 and answer4.numericValue < 0.0721
"You're close — 0.072 is the per-hour growth rate (about 7.2%), not the per-hour factor. The factor includes the original amount plus the growth: 1 + 0.0718 ≈ 1.0718."
when answer4.submitted and not isBlank(answer4.latex)
"I see what happened — what number, multiplied by itself 10 times, gives 2? That is the hourly factor."
otherwise ""
Activity 2 · SynthesisOptional5 min
Bacteria Growth — Discussion
Same function, two different equations — each one highlights something the other hides. Make the trade-off explicit.
“Why are Expression A and Expression B equal?”By the exponent rule \(b^{m \cdot n} = (b^{m})^{n}\) with \(b = 2\), \(m = 1/10\), and \(n = t\):
\[2^{t/10} \;=\; 2^{(1/10)\cdot t} \;=\; \bigl(2^{1/10}\bigr)^{t}\]
Both produce the same value for every t.
“What does each expression highlight?”Expression A \(1000 \cdot 2^{t/10}\) — highlights the doubling time: every 10 hours, multiply by 2. Expression B \(1000 \cdot \!\left(2^{1/10}\right)^{t}\) — highlights the per-hour factor: every 1 hour, multiply by \(2^{1/10} \approx 1.0718\).
“Which would you choose for \(t = 30\)? For \(t = 3\)?”For \(t = 30\): Expression A is cleaner — \(1000 \cdot 2^{30/10} = 1000 \cdot 2^{3} = 8000\). For \(t = 3\): Expression B reads more naturally — the population is multiplied by the hourly factor 3 times. Same function, two equations — pick the one that matches the question being asked.
The discussion's payoff: an exponential equation can be rewritten to surface different rates — per-hour, per-day, per-decade. This is the same property that powered Lesson 5's fractional-interval factor work, but applied in reverse: there, an interval of length \(n\) had factor \(b^{n}\); here, the per-unit factor is rewritten as \(b^{1/n}\) raised to the tth power.
Math As A Second LanguageOptional — pairs with Activity 2
“one thousand times two to the t over ten equals one thousand times the quantity two-to-the-one-tenth, raised to the t”or“the doubling form equals the hourly-factor form”
● Meaning
The same exponential function can be written so the doubling time is visible (left) or so the per-unit factor is visible (right). The exponent rule \(b^{m \cdot n} = (b^{m})^{n}\) connects them.
Optional MASL slide — only show if Activity 2 was completed. The check at \(t = 10\) is the most concrete way to see that “ten hours” and “ten compoundings of the hourly factor” are the same operation. Do not introduce \(2^{1/10}\) as a new fact — students built it themselves on the previous slide.
Lesson Synthesis~5 min
From two points to an equation
A function \(f(x) = a \cdot b^{x}\) has a graph passing through \((0, 300)\) and \((1.5, 75)\), where x is in hours.
“Where on the graph do we see a?”\(a\) is the value of \(f\) when \(x = 0\) — the y-intercept. From the graph, \(a = 300\).
“What equation can we write involving the second point?”\(f(1.5) = 300 \cdot b^{1.5} = 75\), so \(b^{1.5} = \tfrac{75}{300} = \tfrac{1}{4}\). The decay factor over a 1.5-hour interval is \(\tfrac{1}{4}\).
“How do we get the per-unit factor b?”Take the \(\tfrac{1}{1.5}\)-power of both sides — equivalently, the \(\tfrac{2}{3}\)-power: \[b = \!\left(\tfrac{1}{4}\right)^{1/1.5} = \!\left(\tfrac{1}{4}\right)^{2/3} \approx 0.397\]
Check: \(0.397^{1.5} \approx 0.25 = \tfrac{1}{4}\). ✓
“What is the equation?”\(f(x) = 300 \cdot \!\left(\tfrac{1}{4}\right)^{2x/3}\), or equivalently \(f(x) \approx 300 \cdot (0.397)^{x}\). Both forms describe the same function — the first highlights the 1.5-hour quartering, the second highlights the per-hour factor.
click to advance discussion ▶
Try Saying
To write the equation, I first found ___ from the y-intercept, then ___ from the second point.
I Can… (SWBAT)
Write an equation \(f(x) = a \cdot b^{x}\) for an exponential function from two points on its graph — even when the inputs of the points are not 1 unit apart.
Decide what information I need to write an exponential equation, and ask precise questions to get it.
Lesson Synthesis (~5 min). Cold-call through the four reveal points. The conceptual core is question 3 — raising a factor over 1.5 hours to the \(\tfrac{2}{3}\) power gives the per-hour factor. Some students will work in fractional exponents (\((\tfrac{1}{4})^{2/3}\)); others will use a calculator on \(0.25^{1/1.5}\). Both are correct.
SWBAT lives here, not on the title slide. Read aloud once the reveal queue completes.
Cool-Down5 min
Finding a and b
Here is a graph representing an exponential function of the form \(f(x) = a \cdot b^{x}\).
On your own — show your reasoning
Find the value of a.
Find the value of b.
Cool-Down · 5 min. The graph passes through \((0, 8)\) and \((2, 18)\). Find \(a\) and \(b\).
Problem 1: \(a = 8\) (the \(y\)-intercept). Problem 2: \(8 \cdot b^{2} = 18\), so \(b^{2} = \tfrac{18}{8} = \tfrac{9}{4} = 2.25\), so \(b = \sqrt{2.25} = 1.5\).
Common errors: using \(b = 18\) (the \(y\)-value at \(x = 2\), not the per-unit factor); using \(b = 2.25\) (the 2-step factor, not the per-unit factor); using \(b = (18 - 8)/2 = 5\) (linear thinking — the slope, not the factor).
Amplify · Cool-Down Setup
Problem 1: Text component (full problem text + Example: 8) → Math Response named answer1 → Note with CL.
Problem 2: Text component (full problem text + Example: 1.5) → Math Response named answer2 → Note with CL.
Use numericValue ranges for real-number checks.
Problem 1: Find the value of a. The graph passes through (0, 8) and (2, 18).
content:
when answer1.submitted and answer1.numericValue > 7.995 and answer1.numericValue < 8.005
"Correct! a is the y-intercept — the value of f when x = 0. From the graph, a = 8."
when answer1.submitted and answer1.numericValue > 17.995 and answer1.numericValue < 18.005
"Almost — 18 is the y-value at x = 2, not at x = 0. The constant a is the y-intercept (the y-value when x = 0). What is f(0)?"
when answer1.submitted and answer1.numericValue > 1.499 and answer1.numericValue < 1.501
"I see what happened — 1.5 is the per-unit factor b, not a. The constant a is the y-intercept of the graph: f(0) = ?"
when answer1.submitted and answer1.numericValue > 2.249 and answer1.numericValue < 2.251
"Hmm... 2.25 is the 2-step factor (the ratio 18 ÷ 8). The constant a is the y-intercept: f(0) = ?"
when answer1.submitted and answer1.numericValue > 9.995 and answer1.numericValue < 10.005
"Not quite — 10 is the change in y from x = 0 to x = 2 (18 − 8). The constant a is the y-value at x = 0, not the change."
when answer1.submitted and isBlank(answer1.latex)
"Enter the y-intercept of the graph. Example: 8"
when answer1.submitted
"You're close — read the y-coordinate of the point where x = 0. That value is a."
otherwise ""
Problem 2: Find the value of b.
content:
when answer2.submitted and answer2.numericValue > 1.499 and answer2.numericValue < 1.501
"Correct! 8 · b^2 = 18, so b^2 = 18/8 = 2.25, so b = √2.25 = 1.5."
when answer2.submitted and answer2.numericValue > 2.249 and answer2.numericValue < 2.251
"You're close — 2.25 is the 2-step factor (the ratio 18 ÷ 8). The per-unit factor b satisfies b^2 = 2.25, so b = √2.25. What is √2.25?"
when answer2.submitted and answer2.numericValue > 17.995 and answer2.numericValue < 18.005
"Hmm — 18 is the y-value at x = 2, not the per-unit factor. The factor b satisfies 8 · b^2 = 18. Solve for b."
when answer2.submitted and answer2.numericValue > 4.995 and answer2.numericValue < 5.005
"I see what happened — 5 is the average rate of change ((18 − 8) ÷ 2). For exponential functions, the per-unit factor is multiplicative, not additive: 8 · b^2 = 18 gives b = √(18/8) = 1.5."
when answer2.submitted and answer2.numericValue > 7.995 and answer2.numericValue < 8.005
"Not quite — 8 is the y-intercept a, not the per-unit factor b. Use 8 · b^2 = 18 to solve for b."
when answer2.submitted and answer2.numericValue > 1.124 and answer2.numericValue < 1.126
"Good start — 1.125 looks like 18/8/2 (linear logic on the ratio). The per-unit factor is the square root of the 2-step ratio: √(18/8) = √2.25 = 1.5."
when answer2.submitted and answer2.numericValue > 3.354 and answer2.numericValue < 3.356
"Almost — 3.354 looks like √(18 − 8) (taking the root of the difference). The per-unit factor satisfies b^2 = 18/8 = 2.25, so b = √2.25 = 1.5."
when answer2.submitted and isBlank(answer2.latex)
"Enter the per-unit growth factor. Example: 1.5"
when answer2.submitted
"Take another look — use 8 · b^2 = 18, so b^2 = 18/8 = 2.25. Then b = √2.25. What do you get?"
otherwise ""
Lesson Summary
Two points are enough — if you know the distance between them
To write an equation \(f(x) = a \cdot b^{x}\) for an exponential function, we need two pieces of information — usually two points on the graph.
Step 1 — Find a
\(a\) is the y-intercept — the value of \(f\) when \(x = 0\). If the graph passes through \((0, 300)\), then \(a = 300\).
Step 2 — Use the second point
Given \((1.5, 75)\): \(300 \cdot b^{1.5} = 75\), so \(b^{1.5} = \tfrac{1}{4}\).
Step 3 — Solve for b
Raise both sides to the \(\tfrac{1}{1.5} = \tfrac{2}{3}\) power: \(b = \!\left(\tfrac{1}{4}\right)^{2/3} \approx 0.397\).
Lesson Summary
Two points on an exponential graph — together with the distance between their inputs — are enough to write the equation. The per-unit factor is \(b = \!\left(\dfrac{y_{2}}{y_{1}}\right)^{1/n}\), where n is the input distance.
Lesson Summary slide. Read aloud or have a student read. The yellow box generalizes: the per-unit factor is the n-th root of the ratio of the outputs, where n is the distance between inputs. Tomorrow, students will use this skill to compare exponential functions and write equations from word problems involving doubling times and half-lives.
