Modeling with Functions
Unit 3 of Algebra II. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Creating equations, interpreting and analysing functions, building functions from context, inverses, logarithms and exponential equations, comparing models.
How this unit is tested
Start every problem by identifying what kind of relationship the situation describes: constant rate of change means linear, a repeated percent change means exponential, and area/projectile situations mean quadratic. Write the equation using the structure of the model (slope-intercept form, $A(t)=A_0b^t$, or vertex form) before plugging in numbers, and always state a domain that makes sense in the real-world context, not just the algebraic domain.
When a question asks you to interpret or analyze a function, read the question for exactly what feature is being asked for — domain, range, average rate of change, intercepts, or end behavior — and compute that feature directly from the equation or graph rather than describing the function in general terms. Average rate of change is always a slope calculation between two specific points: $\dfrac{f(b)-f(a)}{b-a}$.
For building functions from context, decide whether the situation changes rules at some threshold (piecewise), whether one quantity feeds into another (composite), or whether each term depends on the previous one (recursive). For inverses, remember that swapping x and y and solving for y undoes what the original function does, and that the inverse is only a function if the original passed the horizontal line test — if not, restrict the domain first.
For logarithms and exponential equations, convert between exponential and log form using $b^y=x \iff \log_b x = y$, and choose your method by matching bases when possible; otherwise take a log of both sides. Always check log solutions against the requirement that the argument of any logarithm must be positive. Finally, when comparing models, don't assume exponential beats linear everywhere — check the specific interval being asked about, since exponential growth only guarantees a long-run advantage.
What you have to know
Linear model
A quantity with constant rate of change m and starting value b is modeled by $y=mx+b$.
Exponential growth/decay model
A quantity changing by a fixed percent rate r each period is modeled by $A(t)=A_0(1+r)^t$, where $1+r$ is the growth factor if r>0 and the decay factor if r<0 (equivalently $1-r$ for decay).
Definition of logarithm
For $b>0$, $b\neq 1$, and $x>0$: $\log_b x = y$ if and only if $b^y=x$.
Logarithm rules
$\log_b(mn)=\log_b m+\log_b n$; $\log_b\left(\frac{m}{n}\right)=\log_b m-\log_b n$; $\log_b(m^k)=k\log_b m$.
Inverse function
$f^{-1}$ satisfies $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$; it exists as a function only if $f$ is one-to-one, and its graph is the reflection of $f$ across the line $y=x$.
14 practice questions
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A moving company charges a flat fee of $80 plus $35 per hour of labor. Write a function C(h) for the total cost of h hours of work, and state a reasonable domain for this context.
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Answer. C(h) = 80 + 35h, with domain h ≥ 0.
The flat fee is the constant term and the hourly rate is the constant rate of change, giving a linear model in slope-intercept form. Since negative hours don't make sense, the domain is restricted to non-negative real numbers. -
A ball is thrown upward so that its height in feet after t seconds is modeled by h(t) = -16t^2 + 96t + 4. Find the maximum height the ball reaches and the time at which it occurs.
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Answer. Maximum height is 148 feet at t = 3 seconds.
The vertex of a downward parabola gives the maximum, located at $t=-\dfrac{b}{2a}=-\dfrac{96}{2(-16)}=3$. Substituting back gives $h(3)=-16(9)+96(3)+4=148$. -
For f(x) = x^2 - 3x + 2, find the average rate of change of f over the interval [1,4].
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Answer. 2
Average rate of change is $\dfrac{f(4)-f(1)}{4-1}$. Since $f(4)=6$ and $f(1)=0$, this equals $\dfrac{6-0}{3}=2$. -
Find the domain and range of f(x) = sqrt(2x - 6) + 1.
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Answer. Domain: x ≥ 3. Range: y ≥ 1.
The expression under the square root must be non-negative, so $2x-6\ge0$ gives $x\ge3$. The square root itself is always at least 0, so adding 1 makes the range $y\ge1$. -
A parking garage charges \$5 for the first hour, then \$2 for each additional hour, up to a maximum charge of \$25. Write a piecewise function C(h) for the cost of parking h hours, where 0 < h ≤ 24.
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Answer. C(h) = 5 for 0 < h ≤ 1; C(h) = 5 + 2(h - 1) for 1 < h ≤ 11; C(h) = 25 for h > 11.
The flat \$5 applies during the first hour, then the rate increases linearly by \$2 per additional hour until the total reaches the \$25 cap, which happens exactly at h = 11 since $5+2(10)=25$. After that the cost stays flat at the maximum. -
The radius of a circular oil spill t hours after it began is r(t) = 3t + 2 (in meters), and the area of a circle with radius r is A(r) = πr^2. Write a function modeling the spill's area as a function of time, and find the area after 4 hours.
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Answer. A(t) = π(3t + 2)^2; A(4) = 196π ≈ 615.75 square meters.
This is a composite function A(r(t)), formed by substituting the radius formula into the area formula. At t = 4, the radius is 3(4)+2 = 14, so the area is $\pi(14)^2=196\pi$. -
Find the inverse of f(x) = (2x - 5)/3, and verify that f(f^{-1}(x)) = x.
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Answer. f^{-1}(x) = (3x + 5)/2.
Swap x and y and solve for y: $x=\frac{2y-5}{3}\Rightarrow3x=2y-5\Rightarrow y=\frac{3x+5}{2}$. Checking, $f\left(\frac{3x+5}{2}\right)=\frac{2\cdot\frac{3x+5}{2}-5}{3}=\frac{3x+5-5}{3}=x$, confirming the inverse. -
Let f(x) = x^2 - 4 for x ≥ 0. Find f^{-1}(x) and state its domain.
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Answer. f^{-1}(x) = sqrt(x + 4), domain x ≥ -4.
Restricting to x ≥ 0 makes f one-to-one, so an inverse function exists. Solving $x=y^2-4$ for the non-negative root gives $y=\sqrt{x+4}$, and the domain of the inverse equals the range of the original function, which is $x\ge-4$. -
Solve for x: log_3(x^2 - 5) = 2. Check for extraneous solutions.
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Answer. x = √14 or x = -√14 (both valid).
Convert to exponential form: $3^2=x^2-5$, so $x^2=14$ and $x=\pm\sqrt{14}$. Checking the domain requirement, $x^2-5=9>0$ for both values, so neither solution is extraneous. -
Solve for x: 4^(3x - 1) = 16^(x + 2).
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Answer. x = 5
Rewrite 16 as $4^2$ so both sides share base 4: $4^{3x-1}=4^{2(x+2)}=4^{2x+4}$. Since the bases match, the exponents must be equal: $3x-1=2x+4$, giving $x=5$. -
Solve for x, rounding to three decimal places: 7^x = 250.
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Answer. x ≈ 2.837
Since the bases don't match, take the natural log of both sides: $x\ln7=\ln250$, so $x=\dfrac{\ln250}{\ln7}\approx\dfrac{5.5215}{1.9459}\approx2.837$. -
Two investment options are offered: Option A grows linearly, adding \$500 each year to an initial \$2000. Option B grows exponentially, increasing by 8% each year from an initial \$2000. Write both models and determine which option yields more money after 20 years.
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Answer. A(t) = 2000 + 500t gives A(20) = $12,000. B(t) = 2000(1.08)^t gives B(20) ≈ $9,322. Option A is larger after 20 years.
$1.08^{20}\approx4.661$, so $B(20)\approx2000(4.661)\approx9322$, while $A(20)=2000+500(20)=12000$. This shows that despite exponential growth eventually overtaking linear growth, at t = 20 the linear model is still ahead — the crossover hasn't happened yet. -
A data set gives f(0)=3, f(1)=7, f(2)=15, f(3)=31, f(4)=63. Determine whether this data is best modeled by a linear, quadratic, or exponential function, and justify your answer.
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Answer. Exponential.
The first differences are 4, 8, 16, 32 — not constant, so it's not linear, and the second differences (4, 8, 16) aren't constant either, so it's not quadratic. Since each first difference is exactly double the previous one, the data grows by a constant multiplicative factor, which is the signature of exponential growth. -
A radioactive substance has a half-life of 8 days. If a sample starts with 200 grams, write an exponential decay function for the amount remaining after t days, and determine how many days it takes for the sample to decay to 25 grams.
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Answer. N(t) = 200(1/2)^(t/8); it takes 24 days.
The half-life model is $N(t)=200\left(\frac{1}{2}\right)^{t/8}$. Setting $200(0.5)^{t/8}=25$ gives $(0.5)^{t/8}=\frac{1}{8}=(0.5)^3$, so $\frac{t}{8}=3$ and $t=24$ days.
What people get wrong
- Writing a domain that matches the algebra but not the real situation (e.g., allowing negative time or negative quantity). Instead, always restrict the domain to values that make sense in context, such as $t\ge0$.
- Making an arithmetic slip while isolating y when finding an inverse, then not catching it. Instead, verify every inverse by composing $f(f^{-1}(x))$ and confirming it simplifies to x.
- Treating $\log(a+b)$ as $\log a+\log b$. The addition rule for logs only applies to multiplication inside the log, i.e., $\log(ab)=\log a+\log b$; there is no rule for the log of a sum.
- Accepting every algebraic solution to a log equation without checking that each log's argument is positive. A root that makes an argument zero or negative must be rejected as extraneous.
- Computing average rate of change with mismatched order, e.g. $\dfrac{f(a)-f(b)}{b-a}$. The numerator and denominator must use the same order of points: $\dfrac{f(b)-f(a)}{b-a}$.
- Assuming an exponential model is always larger than a linear model being compared. Growth factors close to 1 with a large linear rate can keep the linear model ahead for a long time before the exponential model overtakes it — check the actual interval asked about.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra II. Free, and no account needed to start.