Selecting a Function Model - Algebra 3

Which function family best models time $t$ as a function of the number of volunteers $n$ ?
Answer: Rational: $t=\frac{36}{n}$ , because time and number of volunteers are inversely proportional. Whenever you see a phrase like "the product of volunteers and time is constant," your first instinct should be inverse variation. That wording is a direct signal for a rational model of the form $t=\frac{k}{n}$ . Here, the constant is $12 \times 3 = 36$ , so the time as a function of volunteers is exactly $t=\frac{36}{n}.$ As $n$ increases, $t$ decreases proportionally: doubling volunteers halves the time, which is the signature of an inverse relationship. The Linear choice $t=36-n$ fails because it claims each extra volunteer subtracts the same number of hours. Actually, the time saved by adding one volunteer depends on how many volunteers you already have: going from 1 to 2 volunteers saves 18 hours, while going from 12 to 13 saves only about 0.23 hours. The Exponential choice $t=36(0.5)^n$ also misses the point: at $n=12$ , it gives $36(0.5)^{12}$ , not 3, so it doesn't even fit the given pair. Exponential functions change by a constant multiplier per added volunteer, not by inverse proportion. The Logarithmic choice $t=36\log(1/n)$ would give negative times for $n>1$ , which is impossible, and logarithms describe slow growth, not inverse variation. On test day, when you see "product is constant" or "inversely proportional," write $xy=k$ immediately. Then solve for the missing variable. That pattern will carry you straight to the rational function.
Which function family best models period $P$ as a function of length $L$ ?
Answer: Power, because $P=kL^{1/2}$ , so the period is a constant multiple of $\sqrt{L}$ . When a question describes a relationship in words like "proportional to the square root of," you should translate that directly into a function equation. Proportional means "constant multiple," so $P$ is proportional to $L^{1/2}$ means $P = kL^{1/2}$ . This is a power function, because the independent variable $L$ is raised to a constant exponent. Check the given data: if $P = k\sqrt{L}$ , then using $P=2.0$ when $L=1$ gives $k=2.0$ . For $L=4$ , $P=2.0\sqrt{4}=4.0$ , exactly matching the passage. So the correct model is the Power choice. Why the others fail: Linear would require $P = mL$ , so doubling length doubles period; here doubling length from 1 m to 2 m should give about $2.0\sqrt{2}\approx 2.83$ s, not 4.0 s. Exponential would require $P = ab^L$ , where equal length increases multiply the period by a constant factor; the data do not show that, and the square-root relationship is not exponential. Quadratic would be $P = kL^2$ , so with $k=2.0$ from the first pair, $L=4$ would give $P=32$ s, not 4.0 s; it confuses squaring with square rooting. Your study takeaway: when you see "proportional to" plus a root or power, write $y = kx^n$ and verify with the given values. That instantly identifies the function family.
Which function family best models total revenue $R$ as a function of ticket price $p$ ?
Answer: Quadratic, because revenue is price times attendance, and attendance is linear in price. Whenever you see a question about revenue, remember the basic structure: revenue is price times quantity sold. The key is to determine what happens to the quantity when price changes. Here, the attendance is described as linear in price: for each $1 increase, 40 fewer people attend. If $p$ is the ticket price, attendance can be written as $A(p)=2000-40(p-50)$ , which simplifies to a linear expression like $A(p)=4000-40p$ . Total revenue is then $R(p)=p\cdot A(p)=p(4000-40p)=-40p^2+4000p$ . Because revenue is the product of price and attendance, and attendance is linear in price, $R$ is quadratic. The choice saying revenue is linear because it is price times a fixed number of tickets misses the fact that attendance is not fixed; it changes with price. The exponential choice misreads the steady loss of 40 attendees as compounding, but the problem gives a constant linear decrease, not a percentage. The rational choice reverses the relationship: revenue is price times attendance, not price divided by attendance, so a rational function does not model this situation. A strong study strategy is to ask yourself: "What changes when the input changes?" If a quantity like attendance is linear in price, and you multiply it by price to get revenue, the result will be quadratic. Watch for that pattern on exam day.
Which function family best models material cost $C$ as a function of radius $r$ ?
Answer: Cubic, because doubling the radius multiplies the volume, and therefore the cost, by 8. When you see a statement like "doubling $r$ makes the cost 8 times as large," your job is to connect that behavior to a function family. This is really about recognizing power functions and proportional reasoning. For a sphere, volume is $V=\frac{4}{3}\pi r^3$ , so if you double $r$ , the volume becomes $\frac{4}{3}\pi (2r)^3 = 8\left(\frac{4}{3}\pi r^3\right)$ . The volume, and therefore the material cost, is multiplied by $8$ . Since cost is proportional to volume, $C = k r^3$ , which is a cubic model. The quadratic choice is a common trap: if cost grew like $r^2$ , doubling $r$ would quadruple the cost, not multiply it by $8$ . The linear choice would mean doubling $r$ doubles the cost, but the test says the factor is $8$ , not $2$ . The exponential choice is also incorrect: exponential functions grow by a constant factor for each equal step in $r$ , but here the factor $8$ is specifically tied to doubling $r$ , and the underlying geometry is a power function, not an exponential one. A useful strategy: whenever a quantity is proportional to a geometric measure like volume, ask what power of the radius is involved. Doubling the radius and observing whether the output multiplies by $2$ , $4$ , or $8$ tells you whether the function is linear, quadratic, or cubic. Here, the factor $8$ points straight to $r^3$ .
A manufacturer records the average cost per unit $C(n)$ for producing $n$ items: $C(100)=8.00$ , $C(200)=5.50$ , $C(400)=4.25$ , $C(800)=3.625$ . The values continue to decrease and appear to approach $3.00 as $n$ grows. Which function family best models $C(n)$ ?
Answer: Rational, because a fixed setup cost is spread over more units: $C(n)=3+\frac{500}{n}$ . The correct model is rational. The values match $C(n)=3+\frac{500}{n}$ : at $n=100$ , $C=8$ ; at $n=200$ , $C=5.5$ ; at $n=400$ , $C=4.25$ ; at $n=800$ , $C=3.625$ . This is the form of a rational function with a horizontal asymptote at $3$ . The exponential option would require a constant percent change over equal steps in $n$ ; the power and logarithmic options do not produce the exact pattern.
Which function family best models the account balance as a function of years since the initial deposit, and why?
Answer: Exponential, because each year the balance is multiplied by the constant factor $1.04$ . Whenever you see a question about how an account balance grows, ask yourself: is the change additive or multiplicative each year? Here, 4% interest compounded annually means the balance is multiplied by $1.04$ every year. After $t$ years, the balance is $B(t)=B_0(1.04)^t,$ where $B_0$ is the initial deposit. That repeated multiplication by a constant factor is the signature of an exponential function, so the correct model is the exponential choice. The linear choice is tempting because interest is paid yearly, but linear growth would require the same dollar amount to be added each year. In reality, 4% of a larger balance is a larger dollar amount, so the yearly increase grows. The quadratic choice also misses the structure: "calculated on an ever-increasing balance" describes compounding, but the pattern is multiplicative, not polynomial. The logarithmic choice gets the direction wrong — logarithmic growth slows over time, but this balance grows faster in dollar terms each year, not slower. For exam day, remember: if a situation says "multiplied by a constant factor" or "increases by a fixed percent," model it exponentially. If it says "increases by a fixed dollar amount," model it linearly. The key is always deciding whether the change is by a constant amount or a constant factor.
Two cars have the same initial value. Car A loses 15% of its current value each year. Car B loses $2,400 each year. Which statement correctly identifies the function families for the values of the cars over time?
Answer: Car A is exponential decay; Car B is linear decay, because Car A's dollar loss decreases and Car B's is constant. The correct model is exponential decay for Car A and linear decay for Car B. A 15% loss of current value multiplies the value by $0.85$ each year, so the dollar loss shrinks over time. A fixed $2,400 loss subtracts the same amount each year, which is linear. The reversed option confuses the two cars; the all-linear and all-exponential options ignore the difference between percent change and constant change.
Which function family best models the machine's value $V$ as a function of time $t$ in years?
Answer: Exponential: $V(t)=20000(0.88)^t$ , because the value is multiplied by 0.88 each year. Whenever a quantity changes by a fixed percent of its current value each period, you are in exponential decay territory. The key signal here is "loses 12% of its current value each year"—the 12% is taken from the value that remains, not from the original $20,000. The correct model is exponential: after one year, the machine retains $100\%-12\%=88\%$ of its value, so its value is multiplied by $0.88$ . After $t$ years, $V(t)=20000(0.88)^t.$ That matches the choice: "Exponential: $V(t)=20000(0.88)^t$ , because the value is multiplied by 0.88 each year." The linear choice, $V(t)=20000-2400t$ , assumes the machine loses exactly $12\%$ of the original $20,000, or $2,400, every year. That would be a fixed dollar amount, but the problem says 12% of the current value, which shrinks each year. The power choice, $20000t^{-0.12}$ , misuses a negative exponent; percent decay is not a power function, and it would not even be defined at $t=0$ . The quadratic choice, $20000(1-0.12t)^2$ , incorrectly treats depreciation as something applied twice rather than as repeated multiplication by 0.88. Study tip: when you see "percent of its current value" or "grows/decays by a percent each year," write the multiplier as $1 \pm r$ and use an exponential function. Fixed dollar change? Then linear. That distinction is the whole question.
Which function family best models perceived loudness $L$ as a function of sound intensity $I$ , and why?
Answer: Logarithmic, because equal multiplicative changes in intensity correspond to equal additive changes in loudness. When you see a statement like "each 10-fold increase in intensity corresponds to an increase of 10 decibels," focus on how the input changes versus how the output changes. Here, the input $I$ changes multiplicatively (multiply by 10), while the output $L$ changes additively (add 10). That pairing—multiplicative input, additive output—is the signature of a logarithmic relationship. So $L = k \log_{10}(I) + c$ describes this exactly: each time $I$ is multiplied by 10, $\log_{10}(I)$ increases by 1, and with a constant $k=10$ , $L$ increases by 10. The linear choice is a trap because it assumes equal additive increases in intensity cause equal increases in loudness, but the passage describes equal multiplicative increases. The exponential choice reverses the pattern: exponential growth means a constant multiplier in the output for a constant addition in the input, which is the opposite of this situation. The power choice misunderstands the decibel scale entirely; if $L$ were proportional to $I^{10}$ , multiplying $I$ by 10 would multiply $L$ by $10^{10}$ , not add 10. As a study tip, whenever a real-world scale says "a 10-fold increase in the original quantity adds a fixed amount to the measured value," think logarithmic. The key is recognizing which variable is multiplied and which is added.
A student makes four claims about function models. Which claim is correct?
Answer: A car's value that decreases by $2,000 each year is linear, because the dollar change is constant. The correct claim is the car with a constant $2,000 yearly loss. A constant dollar change is the signature of a linear model. Constant percent change and constant multipliers are signatures of exponential models, so the population, bouncing ball, and bacteria claims are exponential, not linear.
A student says the decibel level of a sound is a linear function of sound intensity because each time the intensity is multiplied by 10, the decibel level increases by 10. Which statement is correct?
Answer: The student is wrong; the decibel level is a logarithmic function of intensity, so equal multiplicative changes in intensity produce equal additive changes in decibel level. The correct model is logarithmic. The decibel formula is $10\log_{10}(I/I_0)$ , so multiplying $I$ by 10 adds 10 decibels. That is the defining behavior of a logarithmic relationship, not a linear one. The linear option confuses equal multiplicative changes with equal additive changes; the exponential option would have equal additive changes in intensity produce equal multiplicative changes in decibel level; the power option has the wrong structure.
A student computes the third differences of a data set with equally spaced $x$ -values and finds they are all $12$ . The student concludes the data are exponential because the differences are not zero. Which statement is correct?
Answer: The student is wrong; constant third differences indicate a cubic polynomial model. The correct model is cubic. For equally spaced $x$ -values, constant first differences indicate linear, constant second differences indicate quadratic, and constant third differences indicate cubic. Nonzero differences alone do not imply exponential growth; exponential data are identified by a constant multiplicative ratio, not by the size of finite differences.
A student fits a linear model to the height of a ball versus time and then examines the residuals. As time increases, the residuals are negative, then positive, then negative. Which conclusion is most justified?
Answer: The data should be modeled by a quadratic function because the linear fit misses the curvature. The correct model is quadratic. The context is a ball's height over time, which has a curved rise-and-fall shape, and the residual pattern from a linear fit shows that the line misses that curvature. The linear option mistakes a patterned residual plot for randomness; the logarithmic and exponential options do not match a rise-and-fall height model.
A cup of coffee at $90^\circ$ C is placed in a $20^\circ$ C room. Every minute, the temperature difference between the coffee and the room decreases by 8% of its current value. Which function family best models the coffee's temperature $T(t)$ over time?
Answer: Exponential decay: $T(t)=20+70(0.92)^t$ The correct model is exponential decay. The difference $T(t)-20$ is multiplied by $0.92$ each minute, so $T(t)-20=70(0.92)^t$ . The linear option assumes a constant decrease of $5.6^\circ$ per minute; the quadratic option adds a changing rate without the multiplicative decay; the logarithmic option grows without bound and does not match the stated percent decrease.
A quantity $y$ is measured at $x=0,1,2,3,4$ : $5, 7.5, 11.25, 16.875, 25.3125$ . A student says the data cannot be exponential because the increases $2.5, 3.75, 5.625, 8.4375$ are not constant. Which statement is correct?
Answer: The student is wrong; the ratios of consecutive $y$ -values are all $1.5$ , so an exponential model is appropriate. The correct model is exponential. Consecutive ratios are $7.5/5=1.5$ , $11.25/7.5=1.5$ , and so on, so the data are multiplied by $1.5$ at each step. Exponential data do not need constant differences; they need constant multiplicative ratios. The quadratic option is wrong because the second differences are not constant, and the linear option is wrong because a constant percent increase is exponential, not linear.
A ball is thrown upward. Its height $h(t)$ is measured at equally spaced times. The first differences of the height data are $28, 12, -4, -20$ , and the second differences are constant. Which function family best models $h(t)$ ?
Answer: Quadratic, because constant second differences indicate constant acceleration. The correct model is quadratic. Constant second differences are the finite-difference signature of a quadratic function, and constant acceleration produces a quadratic height-time model. The linear option would require constant first differences; the exponential option cannot rise and then fall; the cubic option would require constant third differences, not just changing first differences.
A researcher plots $(\ln x,\ln y)$ for a data set with $x>0$ and finds the points lie on a line with slope $3$ and intercept $-1$ . Which function family is indicated?
Answer: Power: $y=e^{-1}x^3$ The correct model is power. A linear plot of $\ln y$ against $\ln x$ means $\ln y=-1+3\ln x$ , so $y=e^{-1}x^3$ . The exponential option would follow from a linear plot of $\ln y$ against $x$ ; the logarithmic option confuses the variables; the quadratic option has the wrong structure.
A phone data plan costs $40 per month for the first 5 GB. After 5 GB, each additional GB costs $15. A student says the monthly cost is linear because each additional GB has a constant price. Which statement best evaluates this claim?
Answer: The claim is incorrect; the cost is piecewise linear, not a single linear function. The correct model is piecewise linear. For $0\le g\le5$ , the slope is $0 per GB; for $g>5$ , the slope is $15 per GB. A single linear function has one constant slope, so the plan cannot be modeled by one linear equation. The exponential and quadratic options confuse a change in slope with a change in curvature.
Which function family best models the number of views as a function of time since posting?
Answer: Logistic, because the growth is S-shaped and has an upper limit. Whenever you see a question about modeling growth, start by identifying the overall shape and whether there is a limit. The passage describes views that increase slowly, then very rapidly, then slow down as they approach about 50,000 views. That pattern—an S-shaped curve with a horizontal upper bound—is the signature of logistic growth. A logistic model begins with a slow rise, enters a phase of rapid increase, then levels off near a carrying capacity. Here, the 50,000-view ceiling is exactly that capacity, so logistic is the best match. Exponential growth is tempting because of the rapid middle, but exponential functions grow without limit and never slow down, so it cannot account for the leveling off. Logarithmic growth does slow after an initial increase, but it is concave downward from the start and never has the explosive middle stage described. Quadratic growth changes its rate at a constant pace—it can speed up or slow down, but not both—and it has no fixed upper limit. Only logistic combines an initially slow rise, a rapid surge, and a final plateau at a maximum. On exam day, whenever you see phrases like "approaches a maximum" or "levels off," think logistic. If there is no limit, then consider exponential or logarithmic depending on whether the growth rate is increasing or decreasing.
A student records a quantity $y$ at equally spaced $x$ -values and computes the first differences: $4, 6, 8, 10$ . The student claims the data are exponential because the first differences are increasing. Which statement best evaluates this claim?
Answer: The claim is incorrect; the second differences are constant, so a quadratic model is more appropriate. The correct model is quadratic. The first differences $4,6,8,10$ increase by $2$ each time, so the second differences are constant. Constant second differences are the finite-difference signature of a quadratic function. Exponential growth would require a roughly constant multiplicative ratio between consecutive $y$ -values, not merely increasing differences. The linear option is wrong because first differences would be constant; the logarithmic option is wrong because the increases are getting larger, not smaller.

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