AP Precalculus Unit 2 Notes: Understanding Exponentials Through Patterns, Graphs, and Models

Arithmetic and Geometric Sequences

What a sequence is (and why Unit 2 starts here)

A sequence is an ordered list of numbers—called terms—that follow a rule. In precalculus, sequences matter because they are a “discrete” way to describe patterns of change. Exponential functions are “continuous,” but they often come from the same underlying idea as geometric sequences: repeated multiplication by a constant factor.

When you learn to recognize whether a pattern adds the same amount each step (arithmetic) or multiplies by the same factor each step (geometric), you’re training the exact skill you’ll need to choose and justify an exponential model later.

Arithmetic sequences: constant difference

An arithmetic sequence changes by adding (or subtracting) the same number each step. That constant is the common difference.

If the first term is $a_1$ and the common difference is $d$, then each term is:

$a_{n} = a_1 + (n-1)d$

Why it matters: Arithmetic sequences are the discrete version of linear behavior. If data increases by about the same amount each time period, a linear model is often more appropriate than an exponential one.

How it works: You check consecutive differences:

• Compute $a_2-a_1$, $a_3-a_2$, $a_4-a_3$, etc.
• If they’re (approximately) the same, the pattern is arithmetic.

Example 1: Identify and write an explicit formula

Sequence: $5, 9, 13, 17, ...$

The differences are:

• $9-5 = 4$
• $13-9 = 4$
• $17-13 = 4$

So $d = 4$ and $a_1 = 5$. The explicit formula is:

$a_n = 5 + (n-1)4$

You can simplify, but keeping it in this form reduces mistakes.

Geometric sequences: constant ratio

A geometric sequence changes by multiplying by the same number each step. That constant is the common ratio.

If the first term is $a_1$ and the common ratio is $r$, then:

$a_{n} = a_1 r^{n-1}$

Why it matters: This is the discrete version of exponential behavior. If a quantity grows by a constant percent each time period, it forms (approximately) a geometric sequence.

How it works: You check consecutive ratios:

• Compute $\frac{a_2}{a_1}$, $\frac{a_3}{a_2}$, $\frac{a_4}{a_3}$, etc.
• If they’re (approximately) the same, the pattern is geometric.

Example 2: Identify and write an explicit formula

Sequence: $3, 6, 12, 24, ...$

Ratios:

• $\frac{6}{3} = 2$
• $\frac{12}{6} = 2$
• $\frac{24}{12} = 2$

So $r = 2$ and $a_1 = 3$. The explicit formula is:

$a_n = 3 \cdot 2^{n-1}$

Arithmetic vs. geometric (how to tell quickly)

A reliable way to decide is to ask: “Is the change additive or multiplicative?”

A common misconception is thinking “if it increases fast, it must be exponential.” Speed alone doesn’t decide it; the type of change does. For example, adding 100 each step can look “fast,” but it’s still arithmetic.

Exam Focus

• Typical question patterns:
  • Given a list or table, decide whether it’s arithmetic or geometric and write an explicit formula.
  • Find a missing term using a common difference or common ratio.
  • Connect a geometric sequence to an exponential function form.
• Common mistakes:
  • Using differences when you should use ratios (or vice versa). If the data involves percent change, ratios are usually the right tool.
  • Mixing up indexing: writing $a_n = a_1 r^n$ instead of $a_1 r^{n-1}$.
  • Computing ratios when terms can be zero or change sign without checking whether the ratio is consistent.

Exponential Functions and Their Graphs

What an exponential function is

An exponential function is a function where the input appears in the exponent. The most common form in precalculus is:

$f(x) = a b^x$

Here:

• $a$ is the **initial value** (specifically, $f(0)=a$).
• $b$ is the base (the constant multiplicative factor).

Why it matters: Exponential functions model situations with constant multiplicative change—like constant percent growth/decay, compound interest, population growth under ideal conditions, depreciation, and radioactive decay.

How the parameters $a$ and $b$ affect the graph

Think of $b$ as the “multiplier per 1 unit of $x$.”

• If $b > 1$, the function shows exponential growth.
• If $0 < b < 1$, the function shows exponential decay.
• The value $a$ vertically scales the function and sets the y-intercept.

Key graphical features:

• Y-intercept:

$f(0) = a b^0 = a$

• Horizontal asymptote (basic form): For $f(x)=ab^x$ with $b>0$, the graph approaches $y=0$ as $x$ goes in one direction.
• Domain: All real numbers.
• Range: Depends on the sign of $a$. If $a>0$, then $f(x)>0$ for all $x$.

A frequent misconception is mixing up “initial value” with “value at $x=1$.” In the form $ab^x$, the initial value corresponds to $x=0$.

Exponential functions as continuous versions of geometric sequences

A geometric sequence is defined only for integer step numbers $n$:

$a_n = a_1 r^{n-1}$

An exponential function extends that pattern to all real inputs:

$f(x) = a b^x$

If you sample an exponential function at integer values of $x$, you get a geometric pattern in the outputs.

Transformations: shifting and changing the asymptote

Many AP-style problems use a transformed exponential:

$f(x) = a b^{x-h} + k$

Interpretation:

• $h$ shifts the graph horizontally (right if $h>0$).
• $k$ shifts the graph vertically, changing the horizontal asymptote to:

$y = k$

Also, $a$ can reflect the graph across the horizontal asymptote if $a<0$.

Example 1: Identify key features from an equation

Given:

$f(x) = 2 \cdot 3^{x-1} - 5$

• Since $b=3>1$, it’s growth.
• Horizontal shift: right 1.
• Vertical shift: down 5.
• Horizontal asymptote:

$y = -5$

• Value at $x=1$:

$f(1) = 2 \cdot 3^{0} - 5 = 2 - 5 = -3$

Notice how easy it is to misread the “starting value.” The y-intercept is $f(0)$, not the coefficient 2.

Finding an exponential function from points

If you know two points on $f(x)=ab^x$, you can solve for $a$ and $b$.

Suppose you know $f(x_1)=y_1$ and $f(x_2)=y_2$.

Then:

$y_1 = a b^{x_1}$

$y_2 = a b^{x_2}$

Divide the equations to eliminate $a$:

$\frac{y_2}{y_1} = b^{x_2-x_1}$

So:

$b = \left(\frac{y_2}{y_1}\right)^{\frac{1}{x_2-x_1}}$

Then substitute back to find $a$.

Example 2: Build the function from two points

Find $f(x)=ab^x$ if $f(0)=4$ and $f(3)=32$.

From $f(0)=4$:

$4 = a b^0 = a$

So $a=4$.

Use $f(3)=32$:

$32 = 4 b^3$

$8 = b^3$

$b = 2$

So:

$f(x)=4\cdot 2^x$

Exam Focus

• Typical question patterns:
  • Given an equation or graph, identify growth vs decay, y-intercept, and horizontal asymptote.
  • Write an exponential function from two points or from a table (often with equally spaced inputs).
  • Compare how changing $a$, $b$, $h$, or $k$ changes the graph.
• Common mistakes:
  • Treating $b$ as an additive rate instead of a multiplicative factor.
  • Confusing the asymptote $y=k$ with an intercept; the graph may never actually reach $y=k$.
  • Plugging points into $ab^x$ when the given model is really $ab^{x-h}+k$, which changes how you solve.

Exponential Growth and Decay Models

What “growth” and “decay” mean in modeling

In modeling, exponential growth means the quantity increases by a constant percentage per unit time (or per unit input). Exponential decay means it decreases by a constant percentage.

That “constant percent” idea is the core. If something grows by 5% per year, each year it becomes 1.05 times as large as the previous year. That repeated multiplication is exactly exponential behavior.

Two common model forms (and how they connect)

You will see exponential models written in at least two equivalent ways.

Form A: multiplier-per-unit form

$A(t) = A_0 b^t$

• $A_0$ is the initial amount at $t=0$.
• $b$ is the growth factor per 1 unit of $t$.

If the percent growth rate per unit is $r$ (as a decimal), then:

$b = 1 + r$

For decay at rate $r$:

$b = 1 - r$

Form B: percent-change-per-unit form

$A(t) = A_0 (1+r)^t$

This is the same as Form A; it just makes the “percent change” more explicit.

Interpreting growth factors and percent rates

Students often mix up $r$ and $b$. A good habit is to translate in words:

• “Grows by 7% per year” means multiply by 1.07 each year.
• “Decreases by 12% per month” means multiply by 0.88 each month.

The percent rate is the change relative to the current amount, not relative to the starting amount.

Doubling time and half-life (conceptual understanding)

Two important ideas in exponential contexts:

• Doubling time: how long it takes for a growing quantity to become twice as large.
• Half-life: how long it takes for a decaying quantity to become half as large.

In AP Precalculus contexts, you may be asked to estimate these from a table, graph, or repeated-multiplication model. The key is consistency: in an exponential model, equal multiples happen over equal time intervals.

Example 1: Growth model from a percent rate

A bacteria culture starts with 500 cells and grows by 18% per hour. Write a model.

Initial amount:

$A_0 = 500$

Growth factor per hour:

$b = 1 + 0.18 = 1.18$

Model:

$A(t) = 500(1.18)^t$

If you were asked for the number of cells after 6 hours:

$A(6) = 500(1.18)^6$

(You’d typically use a calculator to evaluate.)

Example 2: Decay model from a table with constant percent change

A medication amount is measured every day:

Check ratios:

$\frac{150}{200} = 0.75$

$\frac{112.5}{150} = 0.75$

Constant ratio 0.75 means 25% decay per day. Model:

$A(t) = 200(0.75)^t$

A very common error here is to say “it decays by 50 each day” because the first difference is $-50$. But the next difference is $-37.5$, so it’s not constant additive change.

When your time unit changes

The base (growth factor) depends on the time step. If a quantity grows by 10% per year, the yearly factor is 1.10. That does not mean the monthly factor is 1.10; the monthly factor would need to compound to 1.10 over 12 months.

If the yearly factor is 1.10, and the monthly factor is $m$, then:

$m^{12} = 1.10$

So:

$m = (1.10)^{\frac{1}{12}}$

You don’t always need to compute this exactly, but setting up the relationship correctly is a skill that appears in modeling questions.

Exam Focus

• Typical question patterns:
  • Write an exponential model from an initial value and a percent growth/decay rate.
  • Determine whether data is exponential by checking for an approximately constant ratio (especially with equal time steps).
  • Interpret parameters: explain what $A_0$ and $b$ mean in context.
• Common mistakes:
  • Using $b=r$ instead of $b=1+r$ (for growth) or $b=1-r$ (for decay).
  • Mixing time units (applying a yearly growth factor to monthly steps without adjustment).
  • Deciding “exponential” from one pair of points; you need a consistent multiplicative pattern across intervals.

Competing Function Model Validation

What “model validation” means

In real data, more than one type of function can seem plausible. Competing model validation means you:

1. Propose at least two reasonable models (often linear vs exponential in this unit).
2. Fit or compute each model’s predictions.
3. Compare how well each model matches the data.
4. Justify your choice using evidence, not just a guess.

Why it matters: AP-style questions often ask you to choose between a linear model (constant additive change) and an exponential model (constant multiplicative change). The key skill is defending your choice using patterns in the data and error comparisons.

First-pass checks: difference vs ratio

If the input steps are evenly spaced (for example, every year), two quick diagnostics are:

• Differences (linear check):

$y_{k+1}-y_k$

• Ratios (exponential check):

$\frac{y_{k+1}}{y_k}$

If differences are roughly constant, linear is a strong candidate.
If ratios are roughly constant, exponential is a strong candidate.

This is not a perfect test (real data is noisy), but it’s often the intended reasoning.

Building two models from limited information

Suppose you have data at $t=0$ and $t=n$.

• A linear model through two points has form:

$L(t) = mt + b$

with slope:

$m = \frac{y_2-y_1}{t_2-t_1}$

and intercept $b = y$ at $t=0$ if one of your points is $t=0$.

• An exponential model through two points has form:

$E(t) = A_0 c^t$

If $E(0)=A_0=y_1$, then $A_0$ is fixed, and you solve for $c$ using the second point:

$y_2 = y_1 c^{t_2}$

$c = \left(\frac{y_2}{y_1}\right)^{\frac{1}{t_2}}$

Comparing models using residuals

A residual is:

$\text{residual} = \text{observed} - \text{predicted}$

A model is generally better if residuals are smaller in magnitude and show no obvious pattern (for example, not consistently positive early and negative later).

Even when you’re not asked to make a full residual plot, computing a few residuals is often enough to justify your choice.

Example: Choose between linear and exponential models

Data for a quantity $P$ over time $t$:

Step 1: Quick diagnostic
Differences:

• $65-50 = 15$
• $84-65 = 19$
• $109-84 = 25$

Not constant.

Ratios:

• $\frac{65}{50} = 1.3$
• $\frac{84}{65} \approx 1.2923$
• $\frac{109}{84} \approx 1.2976$

Ratios are fairly close to 1.3, suggesting exponential.

Step 2: Build a linear model using endpoints
Using points $(0,50)$ and $(3,109)$:

$m = \frac{109-50}{3-0} = \frac{59}{3}$

So:

$L(t) = \frac{59}{3}t + 50$

Predictions:

$L(1) = \frac{59}{3} + 50 = \frac{209}{3} \approx 69.67$

$L(2) = \frac{118}{3} + 50 = \frac{268}{3} \approx 89.33$

Residuals:

$\text{residual at } t=1 = 65 - 69.67 \approx -4.67$

$\text{residual at } t=2 = 84 - 89.33 \approx -5.33$

Step 3: Build an exponential model using endpoints
Let:

$E(t) = 50c^t$

Use $t=3$:

$109 = 50c^3$

$c^3 = \frac{109}{50} = 2.18$

$c = (2.18)^{\frac{1}{3}}$

Now predict intermediate values:

$E(1) = 50c$

$E(2) = 50c^2$

(You’d evaluate numerically with a calculator.) If those predictions are closer than the linear predictions—and the ratios were already consistent—your validation supports the exponential model.

What can go wrong: If you only use endpoints, both models will match those endpoints perfectly. Validation comes from checking the other points (and from difference/ratio reasoning).

Using context to validate a model

Sometimes the “best” model is not only about numerical fit but also about whether the mechanism makes sense.

• If the context involves “grows by a constant percent,” “compounds,” or “multiplies,” exponential is mechanistically reasonable.
• If the context involves “adds a constant amount,” “fixed increase per period,” or “constant rate,” linear is more reasonable.

You should be prepared to write a short justification like: “The ratios are approximately constant, indicating a constant percent change, so an exponential model is appropriate.”

Exam Focus

• Typical question patterns:
  • Given a table, decide whether linear or exponential fits better and justify using differences vs ratios.
  • Construct both a linear and exponential model from two points, then compare predictions (often at a middle value).
  • Use residuals (computed or described) to argue which model is more accurate.
• Common mistakes:
  • Choosing a model just because the graph “curves” without checking multiplicative vs additive change.
  • Comparing models only at the points used to build them; you need at least one additional check point.
  • Misinterpreting “percent error” vs “raw error.” When values get large, percent error can be a more meaningful comparison than absolute differences.