The exponential parent function with a base greater than one, the effect of the coefficient a on the y-intercept, translations and horizontal asymptotes, exponential growth models built from a percent increase, and the compound interest formula.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Graph Exponential Growth Functions
Objectives
Five outcomes. The first is a new family of functions; the last two are what it was invented for.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 478-485 — the lesson these objectives are drawn from
Warm-up
Every function family so far has put the variable in the base. This chapter moves it.
Discussion prompt
Compare y equals x cubed with y equals 3 to the x at x equal to 1, 2, 3 and 10. What happens to each as x grows?
Hint: Compute both at 10.
Answer:
\[ x^3: \; 1, 8, 27, 1000 \qquad 3^x: \; 3, 9, 27, 59049 \]
They agree at x equal to 3 and then part company completely. A power function multiplies by a fixed power of the input; an exponential function multiplies by a fixed factor at every step. By x equal to 10 the exponential is nearly sixty times larger, and the gap only widens.
Concept
An exponential function has the form y equals a times b to the x, with a non-zero and the base b positive and not 1. When a is positive and b exceeds 1 the function grows, multiplying by b every time x increases by 1.
exponential growth function — A function y equals a times b to the x with a greater than zero and b greater than 1. The base b is called the growth factor.
\[ y = ab^x, \quad a > 0, \; b > 1 \]
The graph never touches the horizontal axis, however far left it goes, because a positive number raised to any power is positive. That line is an asymptote.
Figure (svg): Two columns contrasting a power function with an exponential function
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 478-478
Section
Section 1
Concept
For a base greater than 1, the function b to the x rises from left to right, passes through the point where x is 0 and y is 1, and through the point where x is 1 and y is b. Its domain is every real number and its range is the positive numbers.
asymptote — A line that a graph approaches more and more closely without ever reaching it. The horizontal axis is an asymptote of the graph of b to the x.
\[ f(x) = b^x, \quad b > 1 \]
Anything to the power zero is 1, which is why every graph of this family crosses at the same height. The base only shows itself from x equal to 1 onward.
Figure (svg): The graph of two to the x with its table of values and its horizontal asymptote
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 478-478 — Parent Function for Exponential Growth Functions
Picture it
Example 1: a table from negative 2 to 3, and the curve through it.
Figure (svg): The graph of two to the x with its table of values and its horizontal asymptote
To the left the values halve at every step — a quarter, a half — and approach zero without reaching it. To the right they double, and the curve climbs steeply.
Worked example
Example 1, in the book's three steps.
\[ \text{Graph } y = 2^x. \]
Make a table of values
Why: At x from negative 2 to 3 the outputs are one quarter, one half, 1, 2, 4 and 8.
Plot them
Why: The points climb steeply to the right and flatten to the left.
Draw the curve from left to right
Why: It begins just above the horizontal axis and rises through the points.
State the domain, range and asymptote
Why: Every real number is an allowed input, and every output is positive.
\[ y > 0,\text{ asymptote } y = 0 \]
Figure (svg): The solution to Worked example graph the parent shown as a ladder of expressions, one row per algebraic move
\[ y = 2^x, \quad y > 0 \]
Verify: check what happens far to the left
Why: At x equal to negative 10 the value is 1 over 1024, which is tiny but positive. At negative 100 it is smaller still and still positive. No input ever makes it zero, which is exactly what having an asymptote means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 478-478
Sorting
Look at where the variable sits.
Sort into buckets
Sort each function.
Two of these use the same two numbers, 2 and x, in opposite positions, and the resulting functions could hardly be more different.
Worked example
Guided Practice 1, plus a comparison.
\[ \text{Graph } y = 4^x \text{ and compare it with } y = 2^x. \]
Tabulate
Why: At x from negative 1 to 2 the outputs are one quarter, 1, 4 and 16.
Find the shared point
Why: Both curves pass through the point where x is 0 and y is 1.
\[ (0, 1)\text{ on both} \]
Compare to the right
Why: At x equal to 1 the second is 4 while the first is 2, and the gap widens.
\[ 4 ^{x}\text{ rises faster} \]
State the domain and range
Why: Unchanged: all reals, and y greater than zero.
Figure (svg): The solution to Worked example two more bases shown as a ladder of expressions, one row per algebraic move
\[ y = 4^x, \quad y > 0 \]
Verify: check the relationship between the two
Why: Four to the x is 2 to the x, squared, so the second curve's height is always the square of the first's. At x equal to 3 that is 8 squared, or 64, and 4 cubed is indeed 64. A larger base is a steeper curve, and every curve in the family shares the point where y is 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479
Trap
\[ y = 2^x \]
Extend the curve down to meet the horizontal axis on the left
Why: The values get very small, so they are taken to reach zero.
\[ 2^x = 0 \text{ for some large negative } x \quad \text{(wrong)} \]
Two to the negative 100 is one over 2 to the hundredth, an extremely small positive number — but positive. No power of 2 is ever zero.
\[ 2^x > 0 \text{ for every real } x \]
Draw the curve approaching the axis without touching it
Why: The horizontal axis is an asymptote, not part of the graph.
\[ \text{range: } y > 0, \text{ never } y = 0 \]
The strict inequality in the range is the whole content of the word asymptote. Getting arbitrarily close and arriving are different things, and only the first happens here.
Fill the middle
Example 1, at x equal to negative 2.
Fill in the blanks
y = 2^x: \; 2^1/4 = \frac______ = ___
Why: A negative exponent gives a reciprocal, as Lesson 5.1 established, so 2 to the negative 2 is one quarter. That is why the left half of the graph sits between zero and 1 and shrinks toward the axis rather than going negative.
Matching
Every graph passes through (1, b).
Match the pairs
Why: All four also pass through the point where x is 0 and y is 1, since anything to the power zero is 1. So one point is shared by the whole family and the next one identifies the base — which makes reading a base off a graph very quick.
Prediction
Commit before reasoning.
Predict first
Why can b to the x never be zero or negative when b is positive?
Correct: Because a positive base raised to any power stays positive.
\[ 2^{-100} = \tfrac{1}{2^{100}} > 0 \]
Why: For a positive exponent the result is a product of positive factors; for a negative exponent it is the reciprocal of such a product, which is still positive; and for exponent zero it is 1. So the output is never zero and never negative, whatever x is. That single fact gives both the range and the asymptote.
Section
Section 2
Concept
The graph of a times b to the x is a vertical stretch or shrink of the graph of b to the x, and its y-intercept is at a rather than 1. A negative a reflects the curve below the horizontal axis.
\[ y = ab^x \;\Longrightarrow\; \text{y-intercept } (0, a) \]
A negative a makes the function decrease, so it is not an exponential growth function even when the base exceeds 1 — the textbook is explicit about this.
Figure (svg): Two exponential functions with different coefficients, one positive and one negative
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479 — Graph y = ab^x for b > 1
Picture it
Example 2: one half times 4 to the x, and negative one times 2.5 to the x.
Figure (svg): Two exponential functions with different coefficients, one positive and one negative
Both curves have the horizontal axis as an asymptote, but one approaches it from above and the other from below. Only the first is a growth function.
Worked example
Example 2, both parts.
\[ \text{Graph } y = \tfrac{1}{2}\cdot 4^x \text{ and } y = -\left(\tfrac{5}{2}\right)^x. \]
First: find two anchor points
Why: At x equal to 0 the value is a, which is one half; at x equal to 1 it is a times b, which is 2.
\[ (0, \frac{1}{2})\text{ and } (1, 2) \]
First: draw the curve
Why: It begins just above the axis on the left and rises to the right.
Second: find two anchor points
Why: At x equal to 0 the value is negative 1; at x equal to 1 it is negative five halves.
\[ (0, -1)\text{ and } (1, -2.5) \]
Second: draw and classify
Why: The curve begins just below the axis and falls to the right; since a is negative it is not a growth function.
Figure (svg): The solution to Worked example two coefficients shown as a ladder of expressions, one row per algebraic move
\[ (0, \tfrac{1}{2}), (1, 2); \qquad (0, -1), (1, -\tfrac{5}{2}) \]
Verify: check the two anchor points in general
Why: At x equal to zero the base disappears, leaving a; at x equal to 1 the value is a times b. So plotting a and then a times b locates any curve in this family in two evaluations, whatever the sign of a. In the second, negative one times 2.5 is negative 2.5, matching the picture.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479
Fill the middle
Example 2a.
Fill in the blanks
y = \tfrac1/2___\cdot 4^x: \; \text___ x = 0, \; y = \tfrac______\cdot 4^0 = \tfrac______\cdot 1 = ___
Why: Four to the power zero is 1, so the value is just a, which is one half. Every function in this family has its y-intercept at a, and reading it off the equation takes no calculation at all.
Worked example
Guided Practice 2, plus one for contrast.
\[ \text{Graph } y = \tfrac{1}{2}\cdot 3^x \text{ and } y = 5\cdot 2^x. \]
First: anchor points
Why: At x equal to 0 the value is one half; at x equal to 1 it is three halves.
\[ (0, \frac{1}{2}), (1, 1.5) \]
First: describe
Why: A vertical shrink of 3 to the x by a factor of one half.
Second: anchor points
Why: At x equal to 0 the value is 5; at x equal to 1 it is 10.
\[ (0, 5), (1, 10) \]
Second: describe
Why: A vertical stretch of 2 to the x by a factor of 5.
Figure (svg): The solution to Worked example two more coefficients shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{2}\cdot 3^x; \qquad y = 5\cdot 2^x \]
Verify: check that a never changes the asymptote
Why: Both curves still approach the horizontal axis on the left, because multiplying something that tends to zero by any constant still tends to zero. The coefficient scales the graph vertically but leaves the asymptote where it is — which the next idea changes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479
Error analysis
A student classifies an exponential function.
Annotate
On: \( y = -\left(\tfrac{5}{2}\right)^x: \text{ the base } 2.5 > 1, \text{ so it is an exponential growth function} \)
Two conditions, both required. The textbook flags this exact case in a Classify Functions note beside Example 2.
Sorting
Both a and b have conditions.
Sort into buckets
Sort each function.
Every base here exceeds 1, so the sign of a is doing all the sorting — which is exactly the point the definition makes.
Prediction
Commit before reasoning.
Predict first
The graph of 2 to the x has the horizontal axis as its asymptote. What about 5 times 2 to the x?
Correct: The asymptote is still the horizontal axis.
\[ 5\cdot 2^{-100} = \tfrac{5}{2^{100}} \to 0 \]
Why: Multiplying a quantity that shrinks toward zero by 5 gives a quantity that still shrinks toward zero. The coefficient scales heights, and five times something arbitrarily small is still arbitrarily small. Only adding a constant moves the asymptote, which is what the next idea does with k.
Comparison
Fill the blanks. The letter a does a familiar job.
Comparison matrix
| Family | Effect of a | Effect of a being negative |
|---|---|---|
| Quadratic, 4.2 | stretches or shrinks vertically | opens the parabola downward |
| Radical, 6.5 | stretches or shrinks vertically | reflects across the horizontal axis |
| Exponential, here | stretches or shrinks vertically | reflects below the horizontal axis |
| All three | the y-intercept moves to a | the graph flips |
The coefficient behaves identically in every family, which is why recognising it costs nothing once the first family has been learned.
Section
Section 3
Concept
To graph a times b to the x minus h, plus k, sketch a times b to the x and then shift it h units horizontally and k units vertically. The asymptote moves from the horizontal axis to the line y equals k, and the range becomes everything on one side of it.
\[ y = ab^{x-h}+k \;\Longrightarrow\; \text{asymptote } y = k \]
The domain stays all real numbers, because no translation can make an exponential function undefined anywhere. Only the range changes.
Figure (svg): An exponential function translated right and down, with its asymptote moved to y equals negative three
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479 — Graph y = ab^(x-h) + k
Picture it
Example 3: 4 times 2 to the x minus 1, minus 3.
Figure (svg): An exponential function translated right and down, with its asymptote moved to y equals negative three
The two anchor points moved with the curve, and so did the asymptote — from the horizontal axis down to the line three units below it.
Worked example
Example 3, with domain and range.
\[ \text{Graph } y = 4\cdot 2^{x-1}-3 \text{ and state the domain and range.} \]
Sketch the untranslated version
Why: Four times 2 to the x passes through the points where x is 0 and y is 4, and where x is 1 and y is 8.
Read h and k
Why: The exponent is x minus 1, so h is 1; the constant outside is negative 3.
\[ h = 1, k = -3 \]
Shift the graph
Why: Right 1 and down 3, carrying both anchor points with it.
\[ (1, 1)\text{ and } (2, 5) \]
State the asymptote, domain and range
Why: The asymptote moves to the line y equals negative 3.
Figure (svg): The solution to Worked example translate an exponential function shown as a ladder of expressions, one row per algebraic move
\[ y > -3, \quad \text{asymptote } y = -3 \]
Verify: check a shifted anchor point
Why: The point where x was 0 and y was 4 should move to x equal to 1 and y equal to 1. Substituting x equal to 1 gives 4 times 2 to the zero, minus 3, which is 4 minus 3, or 1. The translation was applied correctly, and the asymptote at negative 3 is where the curve flattens on the left.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479
Matching
The asymptote is the line y equals k.
Match the pairs
Why: Only the constant added outside the power matters; h and a play no part at all in where the asymptote sits. When there is no constant, k is zero and the asymptote is the horizontal axis, as in the third case.
Worked example
Guided Practice 3.
\[ \text{Graph } f(x) = 3^{x+1}+2 \text{ and state the domain and range.} \]
Read h and k
Why: X plus 1 means h is negative 1, and the constant outside is 2.
\[ h = -1, k = 2 \]
Locate the anchor points
Why: Three to the x passes through the points where y is 1 and where y is 3; shifting left 1 and up 2 moves them.
\[ (-1, 3)\text{ and } (0, 5) \]
Find the asymptote
Why: It moves from the horizontal axis up to the line y equals 2.
\[ y = 2 \]
State the domain and range
Why: Every real input is allowed, and every output exceeds 2.
Figure (svg): The solution to Worked example a left-and-up translation shown as a ladder of expressions, one row per algebraic move
\[ y > 2, \quad \text{asymptote } y = 2 \]
Verify: check the value at zero
Why: Substituting x equal to 0 gives 3 to the first, plus 2, which is 5 — matching the shifted anchor point. And as x runs far to the left, 3 to the x plus 1 shrinks toward zero, so the whole expression shrinks toward 2 without reaching it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 479-479
Trap
\[ y = 4\cdot 2^{x-1}-3 \]
State the range as y greater than zero
Why: The parent's range is copied across unchanged.
\[ \text{range } y > 0 \quad \text{(wrong)} \]
At x equal to negative 5 the value is 4 times 2 to the negative 6, minus 3, which is about negative 2.94 — well below zero.
\[ \text{asymptote } y = k = -3 \;\Longrightarrow\; \text{range } y > -3 \]
Move the asymptote by k, then read the range from it
Why: Adding k to every output shifts both the curve and the line it approaches.
\[ 4\cdot 2^{x-1} > 0 \;\Longrightarrow\; 4\cdot 2^{x-1}-3 > -3 \]
The inequality is the proof: the exponential part is always positive, so subtracting 3 always leaves something above negative 3. The range follows from the algebra, not from the picture.
Fill the middle
Guided Practice 3.
Fill in the blanks
f(x) = 3^-1+2: \; x+1 = x-(___) \;\Longrightarrow\; h = -1
Why: The exponent vanishes at x equal to negative 1, so h is negative 1 and the shift is one unit to the left. Asking what makes the exponent zero gives h directly, the same habit as reading h from a vertex form or a radicand.
Sorting
The sign of a and the value of k together decide it.
Sort into buckets
Sort each function by its range.
The line y equals k is the boundary in every case; a says which side of it the graph lives on, exactly as it did for the radical functions of Lesson 6.5.
Prediction
Commit before reasoning.
Predict first
Translations restricted the domain of a square root function. Why not here?
Correct: Because an exponential function is defined for every real input.
\[ b^x \text{ is defined for every real } x \;\Longrightarrow\; \text{domain always all reals} \]
Why: A square root has a domain condition — its radicand must be non-negative — and shifting the graph moves where that condition bites. An exponential function has no such condition: b to the anything is defined for every real exponent when b is positive. With no restriction to move, no translation can create one.
Section
Section 4
Concept
When a quantity increases by the same percent each period, its amount after t periods is the initial amount times the growth factor raised to t, where the growth factor is 1 plus the percent written as a decimal.
growth factor — The base of an exponential growth function. For a percent increase r written as a decimal, the growth factor is 1 plus r.
\[ y = a(1+r)^t \]
The percent increase and the growth factor are different numbers. A 92 percent increase gives a growth factor of 1.92, and using 0.92 would model a decrease instead.
Figure (svg): An exponential growth model for computer security incidents, with two readings marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 480-480 — Exponential growth models
Picture it
Example 4: 2573 incidents in 1996, rising 92 percent a year.
Figure (svg): An exponential growth model for computer security incidents, with two readings marked
Seven years later the model gives about 247,000 incidents, and the graph reaches 125,000 near t equal to 6, which is 2002.
Worked example
Example 4, in the book's three steps.
\[ \text{With } 2573 \text{ incidents in } 1996 \text{ rising } 92 \text{ percent a year, model } n \text{ and estimate } 2003. \]
Identify a and r
Why: The initial amount is 2573 and the percent increase as a decimal is 0.92.
\[ a = 2573, r = 0.92 \]
Write the model
Why: The growth factor is 1 plus 0.92, which is 1.92.
\[ n = 2573(1.92) ^{t} \]
Evaluate at the right t
Why: The year 2003 is 7 years after 1996.
\[ n = 2573(1.92) ^{7} \]
Compute
Why: One point nine two to the seventh is about 96.2, and 2573 times that is about 247,485.
\[ \text{about } 247, 000 \]
Figure (svg): The solution to Worked example build and use a growth model shown as a ladder of expressions, one row per algebraic move
\[ n = 2573(1.92)^t, \quad n(7) \approx 247{,}485 \]
Verify: check the first two years by hand
Why: After one year the count should be 2573 plus 92 percent of it, which is 2573 plus 2367, or 4940 — and the model gives 2573 times 1.92, which is 4940.16. After two years it is 9485. Each year multiplies by 1.92, which is what a fixed percent increase means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 480-480
Matching
Add 1 to the decimal.
Match the pairs
Why: The last one is the clearest check on the rule: a 100 percent increase is a doubling, and the growth factor is 2. Using 1 instead would leave the quantity unchanged, and using 100 would multiply it by a hundred.
Worked example
Guided Practice 4 and 5.
\[ \text{Estimate the year with about } 250{,}000 \text{ incidents, and describe } y = 527(1.39)^x. \]
Use the graph or the model
Why: The value 247,485 at t equal to 7 is already about 250,000.
\[ t = 7 \]
Translate to a year
Why: Seven years after 1996 is 2003.
\[ 2003 \]
Read the second model's initial amount
Why: The coefficient is the value at x equal to zero.
\[ a = 527 \]
Read its growth factor and percent increase
Why: The base 1.39 is the growth factor, so r is 0.39.
\[ 39 \% \]
Figure (svg): The solution to Worked example read a year, and read a model shown as a ladder of expressions, one row per algebraic move
\[ 2003; \quad 527, \; 1.39, \; 39\% \]
Verify: separate the growth factor from the percent
Why: The growth factor 1.39 and the percent increase 39 are different numbers describing the same thing: multiply by 1.39, or add 39 percent. Reporting the growth factor as the percent increase would claim a 139 percent rise, nearly four times too much.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 480-480
Error analysis
A student writes a growth model for a 92 percent annual increase.
Annotate
On: \( n = 2573(0.92)^t \)
Test the model after one step: it should give the original plus 92 percent of it. The textbook flags this in an Avoid Errors note beside Example 4.
Fill the middle
Example 4, at t equal to 7.
Fill in the blanks
n = 2573(1.92)^7 \approx 2573 \times 96.2 \approx 247___485
Why: One point nine two raised to the seventh is about 96.2, so seven years of 92 percent growth multiply the original by nearly a hundred. That is the practical meaning of exponential growth: a factor slightly under 2 applied seven times is not 14 times but nearly 100.
Prediction
Commit before reasoning.
Predict first
At 92 percent growth a year, roughly how many years to multiply by 100?
Correct: About 7 years.
\[ 1.92^7 \approx 96.2 \quad \text{against} \quad 2^7 = 128 \]
Why: Each year multiplies by 1.92, close to doubling, so seven years multiply by roughly 2 to the seventh, which is 128 — and the exact figure is about 96. This is why the incident count went from 2573 to nearly a quarter of a million in the space of the example. Repeated multiplication outruns intuition quickly, which is the single most important thing this chapter teaches.
Comparison
Fill the blanks. Adding against multiplying.
Comparison matrix
| Question | Linear, Lesson 2.4 | Exponential, here |
|---|---|---|
| Each period | adds a fixed amount | multiplies by a fixed factor |
| The model | y = mx + b | y = a(1 + r)^t |
| 2573 growing by 2367 a year | linear: 21,142 after 7 years | not applicable |
| 2573 growing by 92 percent a year | not applicable | about 247,000 after 7 years |
The two agree after one year, since 92 percent of 2573 is 2367. After seven they differ by a factor of more than ten.
Section
Section 5
Concept
Money in an account earning compound interest grows exponentially. With principal P, annual rate r, and n compoundings a year, the amount after t years is P times the quantity 1 plus r over n, raised to the power n times t.
compound interest — Interest paid on the initial principal and on previously earned interest. Interest paid only on the principal is called simple interest.
\[ A = P\left(1+\tfrac{r}{n}\right)^{nt} \]
Each compounding applies a smaller rate more often. Increasing n raises the balance, but by less and less — a fact Lesson 7.3 turns into the number e.
Figure (svg): The compound interest formula with its four inputs, evaluated at two compounding frequencies
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 481-481 — Compound Interest
Picture it
Example 5: 4000 dollars at 2.92 percent for one year.
Figure (svg): The compound interest formula with its four inputs, evaluated at two compounding frequencies
Quarterly compounding gives 4118.09 dollars and daily gives 4118.52 — a difference of 43 cents on 4000 dollars, which is the whole benefit of compounding ninety times more often.
Worked example
Example 5, both parts.
\[ \text{Find the balance on } 4000 \text{ dollars at } 2.92\% \text{ after } 1 \text{ year, compounded quarterly and daily.} \]
Identify the four inputs
Why: P is 4000, r is 0.0292, t is 1, and n is 4 or 365.
Quarterly: divide and multiply
Why: The rate per quarter is 0.0292 over 4, or 0.0073, applied 4 times.
\[ 4000(1.0073) ^{4} \]
Quarterly: evaluate
Why: One point zero zero seven three to the fourth is about 1.02952.
\[ 4118.09\text{ dollars} \]
Daily: repeat with n equal to 365
Why: The rate per day is tiny and applied 365 times.
\[ 4000(1.00008) ^{365} \]
Daily: evaluate
Why: The result is about 4118.52 dollars.
\[ 4118.52\text{ dollars} \]
Figure (svg): The solution to Worked example two compounding frequencies shown as a ladder of expressions, one row per algebraic move
\[ 4118.09; \qquad 4118.52 \]
Verify: compare with simple interest
Why: Simple interest would give 4000 plus 2.92 percent of 4000, which is 4116.80 dollars. Quarterly compounding adds 1.29 dollars over that and daily adds 1.72. Compounding helps, but at this rate and over one year the amounts are small — the effect grows with the rate and the number of years.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 481-481
Fill the middle
Example 5a.
Fill in the blanks
r = 0.0292, \; n = 4 \;\Longrightarrow\; 1 + \tfrac1.0073___ = 1 + 0.0073 = ___
Why: The annual rate is divided among the four quarters, giving 0.73 percent per quarter. The base of the exponential is that per-period growth factor, and the exponent counts periods rather than years.
Worked example
Guided Practice 6.
\[ \text{Find the balance on } 2000 \text{ dollars at } 4\% \text{ compounded daily after } 3 \text{ years.} \]
Identify the inputs
Why: P is 2000, r is 0.04, n is 365 and t is 3.
Compute the rate per period
Why: Point zero four over 365 is about 0.00010959.
Compute the number of periods
Why: Three hundred sixty-five times 3 is 1095.
\[ 1095\text{ periods} \]
Evaluate
Why: Two thousand times 1.00010959 to the 1095th is about 2254.99.
\[ \text{about } 2255\text{ dollars} \]
Figure (svg): The solution to Worked example three years, compounded daily shown as a ladder of expressions, one row per algebraic move
\[ A \approx 2254.99 \text{ dollars} \]
Verify: check against simple interest
Why: Simple interest would give 2000 plus 3 times 80, which is 2240 dollars. Compounding daily adds about 15 dollars over three years. The gap widens with time because the interest earned in year one itself earns interest in years two and three — which is the whole idea of compounding.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 481-481
Trap
\[ A = P\left(1+\tfrac{r}{n}\right)^{nt} \]
Substitute the annual rate and the number of years directly
Why: The 0.0292 and the 1 are used as they stand, ignoring n.
\[ 4000(1.0292)^1 = 4116.80 \quad \text{(that is simple interest)} \]
This is the balance with no compounding at all. Compounding quarterly should give slightly more, because each quarter's interest earns interest afterwards.
\[ 4000\left(1+\tfrac{0.0292}{4}\right)^{4\cdot 1} = 4000(1.0073)^4 \approx 4118.09 \]
Divide the rate by n and multiply the exponent by n
Why: The rate is shared out across the periods, and there are n times as many of them.
\[ \tfrac{r}{n} \text{ per period}, \; nt \text{ periods} \]
The two changes work against each other, and compounding wins narrowly. Doing one without the other gives an answer that is wildly wrong rather than slightly.
Sorting
More frequent compounding gives more, but not much more.
Sort into buckets
For 4000 dollars at 2.92 percent for one year, sort each frequency by its balance.
The jump from annual to quarterly is worth about 1.29 dollars, and going all the way to daily adds only another 43 cents. The benefit of more frequent compounding is real but sharply limited.
Prediction
Commit before reasoning.
Predict first
Quarterly gives 4118.09 and daily gives 4118.52. What if interest were compounded every second?
Correct: The balance would settle just above 4118.52.
\[ \left(1+\tfrac{r}{n}\right)^{n} \to e^{r} \text{ as } n \to \infty \]
Why: Each increase in n adds less than the last, and the sequence of balances converges. The limiting value is about 4118.53 dollars, and the constant governing that limit is the number e — the subject of Lesson 7.3. Compounding continuously is the best possible case, and it beats daily compounding by about a penny.
Ranking
Using the compound interest formula.
Put in order
Why: Steps two and three are the pair most often half-done. Step five's rounding matters in a money problem: the answer is a balance, so it belongs to the nearest cent rather than carried to eight decimal places.
Comparison
Fill the blanks. This family uses a, h and k the same way as the others.
Comparison matrix
| Letter | What it does | What it does not do |
|---|---|---|
| a | sets the y-intercept and stretches vertically | move the asymptote |
| b | sets how fast the curve grows | affect the y-intercept |
| h | shifts the graph horizontally | change the domain |
| k | shifts the graph vertically | change the domain |
Only k moves the asymptote, and nothing moves the domain, which is all real numbers for every member of this family.
Pattern
One routine for graphing, one for modelling.
For compound interest, divide the annual rate by n and multiply the number of years by n. Doing one without the other is the commonest error.
OpenStax Algebra and Trigonometry 2e, §6.1 Exponential Functions §6.1
Check
The parent function.
Check your understanding
What are the range and asymptote of y = 2^x?
Answer: A
Why: A positive base raised to any power is positive but never zero, so the curve approaches the horizontal axis without reaching it.
Check
A translation. Find the asymptote.
Check your understanding
What are the asymptote and range of y = 4(2^(x-1)) - 3?
Answer: A
Why: The constant k = -3 moves the asymptote, and the positive coefficient keeps the graph above it.
Check
A growth model. Watch the base.
Check your understanding
A quantity starts at 2573 and rises 92 percent a year. What is the model?
Answer: A
Why: The growth factor is 1 + r = 1 + 0.92 = 1.92.
Real world
A town's population is 18,000 and is growing by 3.5 percent a year. A rival town has 25,000 people and is growing by 1.2 percent a year.
Discussion prompt
Write both models, find both populations after 10 years, and estimate when the first town overtakes the second.
Hint: Both are exponential growth models with different growth factors.
Answer:
\[ P_1 = 18000(1.035)^t, \qquad P_2 = 25000(1.012)^t \]
\[ P_1(10) = 18000(1.4106) \approx 25{,}390, \qquad P_2(10) = 25000(1.1268) \approx 28{,}170 \]
After 10 years the towns hold about 25,400 and 28,200 people. Graphing both shows the first overtaking the second after about 15 years, at a population near 30,000.
Two things are worth noticing. The smaller town starts 7000 behind and closes the gap entirely, because a larger growth factor eventually beats any head start — the graphs must cross, whatever the starting values. And the crossing point cannot be found exactly with the tools of this lesson: solving 18000 times 1.035 to the t equals 25000 times 1.012 to the t needs logarithms, which is what Lessons 7.4 to 7.6 are for.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is a 92 percent annual increase modelled by a base of 0.92 or 1.92?
Correct: 1.92 — the new amount is the old plus 92 percent of it.
\[ 1 + r = 1 + 0.92 = 1.92 \]
Why: Growing by 92 percent means the new value is 100 percent plus 92 percent of the old, which is 1.92 times it. A base of 0.92 would leave only 92 percent of the previous value, a decrease of 8 percent a year. Testing one step settles it: 2573 grown by 92 percent is 4940, and 2573 times 1.92 is 4940.16 while 2573 times 0.92 is 2367. The textbook flags this in an Avoid Errors note.
Explain it
They can graph power functions and have just been shown 2 to the x.
Discussion prompt
In four sentences or fewer, explain how an exponential function differs from a power function and what its graph looks like.
Hint: Talk about where the variable sits.
Answer:
In a power function the variable is the base and the exponent is fixed, so the output is a fixed power of the input. In an exponential function they swap: the base is fixed and the variable is the exponent, so every time the input goes up by 1 the output is multiplied by that fixed base.
The graph therefore climbs faster and faster to the right, and to the left it shrinks toward the horizontal axis without ever touching it. It always passes through the point where the input is zero and the output is 1, because anything to the power zero is 1.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For asymptotes, only k matters: the line is y equals k. For growth factors, add 1 to the decimal and test one step. For h, ask what makes the exponent zero. For compound interest, divide the rate by n and multiply the years by n, both every time. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take y equals 4 times 2 to the x minus 1, minus 3, and build the page around it. Top left: sketch y equals 2 to the x with a table from negative 2 to 3, marking the two anchor points and drawing the asymptote as a dashed line. Top right: sketch y equals 4 times 2 to the x, then the fully translated graph, marking where each anchor point moved to and where the asymptote moved to. Bottom left: state the domain, range and asymptote, with a sentence for each saying which letter of the equation produced it. Bottom right: write a growth model of your own from an initial amount and a percent increase you invent, evaluate it at t equal to 0, 1 and 10, and check that the first two agree with adding the percent by hand. In a margin, write the compound interest formula and label all four inputs.
If your bottom-right check disagrees at t equal to 1, you have almost certainly used the percent rather than one plus the percent as the base.
Recap
Five things, and the family is new even though the transformations are not.
| If you see | Then |
|---|---|
| A variable in the exponent | An exponential function |
| A base greater than 1 with a positive a | Exponential growth |
| A negative a | Not a growth function; the graph is below the axis |
| A constant k added outside | The asymptote is y = k |
| A percent increase r | The growth factor is 1 + r |
| Compounding n times a year | Divide r by n and multiply t by n |
Lesson 7.2 keeps everything about this family and changes one condition: a base between 0 and 1, which turns growth into decay.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.1 Graph Exponential Growth Functions §7.1, pp. 478-485 — everything on these slides traces back here
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