Math I · F-LE.5

Interpreting Parameters in Linear and Exponential Functions in Context

Parameters are the control knobs of a model. Learning to interpret them tells students what a formula is actually saying about starting value, rate, growth factor, percent change, fixed cost, baseline, and long-term behavior.

Concept Functions
Domain Linear, Quadratic, and Exponential Models
Read time 11 minutes

What this learning objective is really asking you to learn

This objective asks students to explain what the numbers in a linear or exponential function mean in the real situation being modeled. A parameter is a number or symbol in a formula that controls the behavior of the function. In \(C(m) = 40 + 12m\), the numbers 40 and 12 are parameters. In \(P(t) = 500(1.08)^t\), the numbers 500 and 1.08 are parameters. Students are not just expected to calculate with these numbers. They are expected to interpret them.

In a linear function, the most common form is \(f(x) = mx + b\). The parameter \(m\) is the slope, or constant rate of change. It tells how much the output changes when the input increases by 1 unit. The parameter \(b\) is the output when the input is 0. Depending on the context, \(b\) might be a starting amount, initial height, base fee, fixed cost, opening balance, y-intercept, or baseline value.

For example, suppose \(C(h) = 25 + 18h\) represents the cost in dollars to rent equipment for \(h\) hours. The 25 means there is a fixed cost of 25 dollars before any hourly time is added. The 18 means the cost increases by 18 dollars for each additional hour. The input unit is hours. The output unit is dollars. The slope unit is dollars per hour. Without units, the interpretation is incomplete.

A negative slope also has meaning. If \(D(t) = 500 - 60t\) represents the distance in miles remaining after \(t\) hours of driving toward a destination, the slope -60 means the distance remaining decreases by 60 miles each hour. The intercept 500 means the trip begins 500 miles away. The negative sign is not a mistake; it tells the direction of change.

In an exponential function, a common form is \(f(x) = a \cdot b^x\). The parameter \(a\) is the value when \(x = 0\), if there is no vertical shift. It often represents an initial amount. The parameter \(b\) is the factor by which the output is multiplied for each 1-unit increase in input. If \(b = 1.06\), the output grows by 6 percent per step. If \(b = 0.82\), the output keeps 82 percent of its previous value each step, which means it decreases by 18 percent per step.

Exponential models are often written as \(f(t) = a(1 + r)^t\) for growth or \(f(t) = a(1 - r)^t\) for decay. In that form, \(r\) is the percent rate written as a decimal. A value of \(r = 0.03\) means 3 percent growth per time unit. A value of \(r = 0.20\) in a decay model means 20 percent decrease per time unit. Students must understand that the factor is not the same as the percent. A 7 percent increase corresponds to the factor 1.07. A 7 percent decrease corresponds to the factor 0.93.

Some exponential functions include a vertical shift, such as \(f(x) = a \cdot b^x + k\). The parameter \(k\) moves the graph up or down and often represents a baseline, floor, ceiling, environmental level, or long-term limiting value. For example, if \(T(t) = 68 + 22(0.8)^t\) models the temperature of a hot drink in a 68-degree room, then 68 is the room temperature the drink approaches over time. The 22 is the initial difference between the drink temperature and the room temperature if \(t = 0\). The 0.8 means the difference from room temperature keeps 80 percent of its previous value each time unit. This interpretation is more sophisticated than simply saying “68 moves the graph up.” In context, 68 is the surrounding temperature.

This objective is really about meaning. A student who can solve equations but cannot interpret parameters may get correct numerical answers while missing the point of the model. A student who can interpret parameters knows what each number does, why it belongs in the formula, what units it carries, and how changing it would change the situation.

Why students should learn this math

Students should learn parameter interpretation because formulas are used constantly to describe real choices. The numbers in a formula are not decorations. They are claims. A rate in a contract is a claim about how cost changes. A starting value in a model is a claim about the initial condition. A growth factor is a claim about repeated percent change. A baseline is a claim about a long-term level. If students do not know how to interpret parameters, they are vulnerable to misunderstanding the very information formulas are meant to communicate.

This matters in money decisions. A cell phone plan might be modeled by \(C(g) = 30 + 8g\), where \(g\) is gigabytes of extra data. The 30 is the base monthly charge, and the 8 is the cost per extra gigabyte. A loan balance might be modeled by an exponential expression if interest compounds. A savings account might grow according to a factor such as 1.02 per period. Students who can interpret parameters can compare plans and understand what makes one option expensive or affordable.

It matters in work and wages. A pay function might have a starting bonus and an hourly rate. A salary model might include annual percent raises. A commission model might include base pay plus a percentage of sales. The parameters tell where the money comes from: fixed amount, per-unit rate, percent growth, or multiplier.

It matters in science. In a temperature model, a parameter may represent room temperature. In a population model, a parameter may represent initial population or growth factor. In a medication model, a parameter may represent initial dosage or decay factor. In a physics model, a slope may represent speed. In every case, the number has a physical meaning. Interpreting it connects mathematics to the world.

It matters in data literacy. Graphs and formulas appear in news, research summaries, dashboards, and reports. A model might use parameters to describe trends. If a student sees a formula but cannot interpret the parameters, the formula remains opaque. If the student can interpret them, the formula becomes a compressed explanation of the situation.

This objective also helps students learn responsibly. A parameter may have a clear mathematical meaning but an unreasonable contextual interpretation. For example, the y-intercept of a linear model might represent the predicted value at time 0, but time 0 may be outside the domain of the data. A slope may describe average change within a range but not forever. An exponential growth factor may apply for a limited period but not indefinitely. Interpreting parameters forces students to ask whether the meaning is valid in context.

The deeper reason students should learn this objective is that it turns math from symbol pushing into explanation. A student who says “the slope is 12” has done a little math. A student who says “the slope is 12 dollars per month, meaning the bill increases by 12 dollars for each additional month of service” has done useful math. The second student can communicate, defend, and apply the result.

Where this objective fits on the full map of mathematics

On the big map of mathematics, parameters are everywhere. In algebra, parameters define families of equations. In functions, parameters control shape, position, rate, and scale. In statistics, parameters describe populations and models. In geometry, parameters appear in equations for lines, circles, and transformations. In calculus, parameters control motion, growth, decay, and optimization. Objective 033 is an early formal encounter with this idea.

The objective builds directly on seeing structure in expressions. Students have already learned to interpret terms, factors, and coefficients. Now they interpret the main numbers in function models. This is also connected to function transformations. Changing \(b\) in \(mx + b\) shifts a line. Changing \(m\) changes its steepness and direction. Changing \(a\), \(b\), or \(k\) in an exponential function changes starting value, growth factor, and asymptote. The algebraic change and graphical change are connected to contextual meaning.

This objective also prepares students for regression and model fitting in statistics. When technology fits a line to data, it may produce a slope and intercept. Those parameters must be interpreted in context. A slope in a data model is not just a number; it describes an estimated change in output per unit of input. An intercept may or may not be meaningful depending on the situation.

In later math courses, students will interpret parameters in quadratic functions. In \(h(t) = -16t^2 + vt + s\), parameters can represent initial velocity and initial height. In trigonometric functions, parameters represent amplitude, period, phase shift, and midline. In logarithmic and rational functions, parameters affect asymptotes and scale. The habit is the same: identify the number, identify its mathematical job, identify its real meaning.

This objective is also part of mathematical modeling. A model is not complete just because it has a formula. A model must be interpreted. The parameters explain how the situation has been simplified and what assumptions the model is making.

The historical machinery behind parameters

The development of algebra made it possible to write general relationships with symbols. Instead of solving only one numerical problem, mathematicians could describe whole families of problems. Parameters are part of that generality. The formula \(y = mx + b\) does not describe only one line. It describes a family of lines. Choosing \(m\) and \(b\) chooses a specific member of the family.

This shift was powerful in science. A physical law often has constants that must be measured for a particular situation. A model of motion, cooling, growth, or decay may have the same structure across many situations, but different parameter values. The structure tells the type of relationship. The parameters adapt the structure to the real case.

In analytic geometry, parameters allowed mathematicians to classify shapes and transformations. A circle could be described by center and radius. A line could be described by slope and intercept. A parabola could be described by coefficients that affect its direction, width, and position. The same formula type could describe many different objects.

In finance, parameters became essential for interest formulas. Principal, rate, compounding frequency, and time all control the result. In population models, initial population and growth rate control future population. In engineering, parameters describe material properties, dimensions, tolerances, and rates. The modern world runs on models whose parameters must be understood, measured, adjusted, and questioned.

Students learning parameter interpretation are therefore learning a professional habit. Engineers, scientists, economists, programmers, architects, and analysts do not merely calculate outputs. They ask what the parameters mean and whether those values are reasonable.

The technical machinery: interpreting linear parameters

For a linear model \(y = mx + b\), students should interpret \(m\) as the change in output divided by change in input. The units are output units per input unit. If \(y\) is dollars and \(x\) is hours, the slope is dollars per hour. If \(y\) is miles and \(x\) is gallons, the slope is miles per gallon. Units are not extra; they are part of the interpretation.

The intercept \(b\) is the output when the input is 0. In context, it often means the starting amount. But students should be careful. Sometimes input 0 is meaningful: time 0, 0 miles, 0 months, 0 items. Sometimes input 0 is outside the reasonable domain. For example, a model predicting adult height from age may have an intercept that is not meaningful if the data only covers ages 10 to 18. The number still has a mathematical role, but the contextual interpretation may be limited.

A positive slope means the output increases as input increases. A negative slope means the output decreases. A slope of 0 means the output is constant. The larger the absolute value of the slope, the steeper the line and the faster the change per unit.

For example, \(R(t) = 1200 - 75t\) might model remaining money in a travel budget after \(t\) days. The parameter 1200 means the trip begins with 1,200 dollars. The parameter -75 means the budget decreases by 75 dollars per day. The model assumes a constant daily spending rate. That assumption may be reasonable for planning, but actual spending may vary.

The technical machinery: interpreting exponential parameters

For an exponential model \(y = a \cdot b^x\), students should interpret \(a\) as the initial value when \(x = 0\), assuming no vertical shift. The base \(b\) is the multiplier per input unit. If \(b = 1.25\), the output is multiplied by 1.25 each step, meaning a 25 percent increase. If \(b = 0.6\), the output is multiplied by 0.6 each step, meaning a 40 percent decrease.

The percent change can be found from the factor. For growth, percent increase is \((b - 1) \cdot 100%\). For decay, percent decrease is \((1 - b) \cdot 100%\) when \(0 < b < 1\). Students should avoid saying “the growth rate is 1.08” when they mean “the growth factor is 1.08” or “the percent increase is 8 percent.” The distinction matters.

If the model is \(y = a(1 + r)^x\), then \(r\) is the growth rate per unit. If the model is \(y = a(1 - r)^x\), then \(r\) is the decay rate per unit. But this form only works when the rate is expressed as a decimal and the time unit matches the exponent step.

For a shifted exponential model \(y = a \cdot b^x + k\), the parameter \(k\) is the horizontal asymptote for many basic exponential models. It often represents a baseline or limiting value. The initial value is no longer simply \(a\); it is \(a + k\) when \(x = 0\). This is a common source of errors. In \(T(t) = 70 + 50(0.75)^t\), the initial temperature is 120, not 50. The 50 represents the initial amount above the 70-degree baseline.

Worked example: interpreting a subscription model and a depreciation model

Suppose \(C(m) = 14.99 + 2.50m\) gives the monthly cost in dollars for a subscription with \(m\) extra downloads. The parameter 14.99 is the base monthly subscription cost. The parameter 2.50 is the cost per extra download. The slope unit is dollars per download. If the company raises 2.50 to 3.00, the cost becomes more sensitive to downloads. If it raises 14.99 to 19.99, every customer pays more even before extra downloads.

Now suppose \(V(t) = 24000(0.82)^t\) models a car's value in dollars after \(t\) years. The parameter 24,000 is the starting value. The factor 0.82 means the car keeps 82 percent of its value each year, so it loses 18 percent per year under the model. The model assumes the same percent depreciation every year. That may be a simplification, but it gives a clear structure.

The difference between these examples is important. In the subscription model, the cost changes by addition: every extra download adds 2.50 dollars. In the depreciation model, the value changes by multiplication: every year multiplies the current value by 0.82. The parameters reveal the machinery.

Common mistakes and what they reveal

One common mistake is saying that \(b\) in \(a \cdot b^x\) is the amount added each time. It is not. It is the factor multiplied each time. If \(b = 1.1\), the output does not add 1.1 each step; it increases by 10 percent each step.

Another mistake is ignoring units. “The slope is 5” is incomplete. Five what per what? Dollars per mile? Meters per second? Points per game? Units turn a number into meaning.

A third mistake is assuming every intercept is meaningful. The y-intercept has a mathematical meaning, but the real-world interpretation depends on whether input 0 makes sense. A student should say when the interpretation is valid and when it is only a feature of the equation.

A fourth mistake is treating parameters as fixed truths instead of model assumptions. A growth factor may come from data, a contract, or an estimate. If the situation changes, the parameter may change. Models are powerful because they simplify, but simplification should be noticed.

The big takeaway

Parameters are the numbers that make a function model specific. In linear functions, parameters usually tell starting value and constant rate of change. In exponential functions, parameters tell starting value, growth or decay factor, percent change, and sometimes a baseline or limiting value. This objective matters because a formula without interpretation is only half useful. The real power comes when students can say what each number means, what units it has, what assumption it represents, and how it affects the situation.

Problem Library

Problems in the App From This Objective

162 problems across 12 archetypes in the app.

connect slope to rate of change and units.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Interpret the slope 25 in the linear function C(h)=25h+40 for cost in dollars for h hours of repair work.

Problem 2

Interpret the slope -3 in the linear function d(t)=-3t+60 for distance in miles remaining after t hours.

Problem 3

Interpret the slope 2 in the linear function P(g)=2g+10 for points after g games.

Open in simulator
Problem 4

Interpret the slope 0.5 in the linear function T(m)=0.5m+10 for temperature in degrees Celsius after m minutes.

Problem 5

Interpret the slope -2 in the linear function W(d)=50-2d for weight in pounds of a dog after d days on a diet.

Problem 6

Interpret the slope 15 in the linear function A(p)=15p+100 for amount of money in dollars in an account after p payments.

Problem 7

Interpret the slope 0.75 in the linear function H(w)=0.75w+60 for height in inches of a plant after w weeks.

Problem 8

Interpret the slope -500 in the linear function V(t)=10000-500t for value in dollars of a car after t years.

Problem 9

Interpret the slope 0.25 in the linear function E(k)=0.25k+5 for energy consumption in kilowatt-hours for k appliances.

Problem 10

Interpret the slope 12 in the linear function S(h)=12h+200 for sales in units after h hours of advertising.

Problem 11

Interpret the slope -0.1 in the linear function R(x)=-0.1x+10 for remaining fuel in gallons after x miles driven.

Problem 12

Interpret the slope 3 in the linear function B(m)=3m+50 for bacteria count in thousands after m minutes.

connect intercept to initial value.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Interpret the y-intercept 40 in the linear function C(h)=25h+40 for cost in dollars for h hours of repair work.

Problem 14

Interpret the y-intercept 60 in the linear function d(t)=-3t+60 for distance in miles remaining after t hours.

Problem 15

Interpret the y-intercept 12 in the linear function T(n)=4n+12 for tiles in figure n.

Problem 16

Interpret the y-intercept 1500 in the linear function P(t) = 1500 + 50t for population of a town after t years.

Problem 17

Interpret the y-intercept 10 in the linear function H(x) = 0.5x + 10 for height of a plant in cm after x days.

Problem 18

Interpret the y-intercept 200 in the linear function S(w) = 200 - 10w for amount of money in a savings account after w weeks.

Open in simulator
Problem 19

Interpret the y-intercept 15 in the linear function F(g) = 15 - 0.5g for fuel in a car's tank in gallons after g units of distance driven.

Problem 20

Interpret the y-intercept 300 in the linear function W(d) = 300 + 2d for weight of an animal in kg after d days.

Problem 21

Interpret the y-intercept 50 in the linear function B(m) = 50 + 5m for number of bacteria in a culture after m minutes.

Problem 22

Interpret the y-intercept 1000 in the linear function V(s) = 1000 - 20s for volume of water in a tank in liters after s seconds.

Problem 23

Interpret the y-intercept 120 in the linear function R(x) = 120 - 5x for remaining pages to read after x hours.

Problem 24

Interpret the y-intercept 2.50 in the linear function C(t) = 0.75t + 2.50 for cost of a taxi ride in dollars for t miles.

connect leading coefficient to starting amount.
15 problems Warmup Practice Mixed Review Assessment
Problem 25

Interpret the initial value 800 in the exponential function P(t)=800(1.03)^t for population after t years.

Problem 26

Interpret the initial value 1200 in the exponential function A(t)=1200(0.85)^t for value of equipment after t years.

Problem 27

Interpret the initial value 50 in the exponential function M(t)=50(0.5)^t for medicine amount after t hours.

Problem 28

Interpret the initial value 5000 in the exponential function V(t)=5000(1.04)^t for value of an investment after t years.

Problem 29

Interpret the initial value 200 in the exponential function B(h)=200(1.2)^h for number of bacteria after h hours.

Problem 30

Interpret the initial value 100 in the exponential function R(d)=100(0.9)^d for amount of radioactive substance after d days.

Open in simulator
Problem 31

Interpret the initial value 75 in the exponential function T(m)=75(0.95)^m for temperature of a liquid after m minutes.

Problem 32

Interpret the initial value 10000 in the exponential function U(w)=10000(1.08)^w for number of website users after w weeks.

Problem 33

Interpret the initial value 30000 in the exponential function C(y)=30000(0.88)^y for value of a car after y years.

Problem 34

Interpret the initial value 1013 in the exponential function P(h)=1013(0.999)^h for atmospheric pressure at h meters above sea level.

Problem 35

Interpret the initial value 200 in the exponential function D(h)=200(0.75)^h for concentration of a drug in the bloodstream after h hours.

Problem 36

Interpret the initial value 2 in the exponential function H(y)=2(1.15)^y for height of a tree after y years.

Problem 37

Interpret the initial value 1000 in the exponential function I(d)=1000(0.92)^d for light intensity after passing through d meters of water.

Problem 38

Interpret the initial value 2500 in the exponential function S(t)=2500(1.025)^t for amount in a savings account after t years.

Problem 39

Interpret the initial value 5 in the exponential function V(d)=5(1.5)^d for number of people infected with a virus after d days.

connect base to growth or decay factor.
15 problems Warmup Practice Mixed Review Assessment
Problem 40

Interpret the base 1.03 in the exponential function P(t)=800(1.03)^t for population after t years.

Problem 41

Interpret the base 0.85 in the exponential function A(t)=1200(0.85)^t for value after t years.

Problem 42

Interpret the base 2 in the exponential function B(t)=75(2)^t for bacteria count after t hours.

Problem 43

Interpret the base 1.05 in the exponential function V(t)=5000(1.05)^t for investment value after t years.

Problem 44

Interpret the base 0.9 in the exponential function M(h)=200(0.9)^h for milligrams of medication remaining after h hours.

Problem 45

Interpret the base 1.15 in the exponential function A(d)=100(1.15)^d for area in square feet after d days.

Open in simulator
Problem 46

Interpret the base 0.5 in the exponential function R(y)=1000(0.5)^y for grams of radioactive material remaining after y years.

Problem 47

Interpret the base 1.2 in the exponential function S(q)=500(1.2)^q for sales units after q quarters.

Problem 48

Interpret the base 0.88 in the exponential function C(y)=25000(0.88)^y for car value after y years.

Problem 49

Interpret the base 3 in the exponential function N(h)=10(3)^h for number of bacteria after h hours.

Problem 50

Interpret the base 0.7 in the exponential function P(a)=100(0.7)^a for atmospheric pressure at altitude a in km.

Problem 51

Interpret the base 1.008 in the exponential function E(m)=1000(1.008)^m for energy consumption after m months.

Problem 52

Interpret the base 0.995 in the exponential function W(d)=500(0.995)^d for water level in cm after d days.

Problem 53

Interpret the base 1.012 in the exponential function Pop(x)=50000(1.012)^x for population after x decades.

Problem 54

Interpret the base 0.65 in the exponential function I(d)=100(0.65)^d for light intensity in lumens after passing through d filters.

compute percent growth or decay from multiplier.
15 problems Warmup Practice Mixed Review Assessment
Problem 55

Convert the exponential base 1.18 to a percent change.

Problem 56

Convert the exponential base 0.72 to a percent change.

Problem 57

Convert the exponential base 2 to a percent change.

Problem 58

Convert the exponential base 0.995 to a percent change.

Problem 59

Convert the exponential base 1.05 to a percent change.

Problem 60

Convert the exponential base 0.9 to a percent change.

Problem 61

Convert the exponential base 1.5 to a percent change.

Problem 62

Convert the exponential base 0.5 to a percent change.

Problem 63

Convert the exponential base 3 to a percent change.

Problem 64

Convert the exponential base 0.01 to a percent change.

Problem 65

Convert the exponential base 1.001 to a percent change.

Problem 66

Convert the exponential base 0.9999 to a percent change.

Problem 67

Convert the exponential base 1.255 to a percent change.

Open in simulator
Problem 68

Convert the exponential base 0.875 to a percent change.

Problem 69

Convert the exponential base 10 to a percent change.

write multiplier from context.
15 problems Warmup Practice Mixed Review Assessment
Problem 70

Convert the percent change 7% increase to an exponential base.

Open in simulator
Problem 71

Convert the percent change 30% decrease to an exponential base.

Problem 72

Convert the percent change 0.4% increase to an exponential base.

Problem 73

Convert the percent change 12.5% decrease to an exponential base.

Problem 74

Convert the percent change 2% increase to an exponential base.

Problem 75

Convert the percent change 50% increase to an exponential base.

Problem 76

Convert the percent change 1% decrease to an exponential base.

Problem 77

Convert the percent change 90% decrease to an exponential base.

Problem 78

Convert the percent change 100% increase to an exponential base.

Problem 79

Convert the percent change 100% decrease to an exponential base.

Problem 80

Convert the percent change 0.1% decrease to an exponential base.

Problem 81

Convert the percent change 25% increase to an exponential base.

Problem 82

Convert the percent change 75% decrease to an exponential base.

Problem 83

Convert the percent change 150% increase to an exponential base.

Problem 84

Convert the percent change 0% increase to an exponential base.

explain changed slope/intercept.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Interpret the transformed linear model C(h)=30(h-2)+90 in cost after h hours with first 2 hours included in a base charge.

Problem 86

Interpret the transformed linear model F(c)=1.8c+32 in temperature conversion from Celsius to Fahrenheit.

Problem 87

Interpret the transformed linear model d(t)=50-4(t-1) in distance remaining after t hours.

Problem 88

Interpret the transformed linear model P(y) = 500 + 20(y-2000) in population P in year y, where y is years after 2000.

Problem 89

Interpret the transformed linear model V(t) = 1500 + 75t in value of an investment V after t years.

Problem 90

Interpret the transformed linear model R(x) = 100 - 5(x-10) in remaining amount R after x items are sold, starting from 10 items.

Problem 91

Interpret the transformed linear model H(m) = 2000 - 10m in height H of a balloon after m minutes.

Problem 92

Interpret the transformed linear model L = 0.5(w-10) + 20 in length L of a spring when a weight w is attached, starting from w=10.

Problem 93

Interpret the transformed linear model C = 15 + 2.5g in cost C of a taxi ride based on g miles traveled.

Problem 94

Interpret the transformed linear model B(s) = 5000 - 200(s-5) in balance B in a bank account after s withdrawals, starting from 5 withdrawals.

Problem 95

Interpret the transformed linear model W = 100 - 0.25h in amount of water W remaining in a tank after h hours.

Open in simulator
Problem 96

Interpret the transformed linear model S(m) = 1/2(m-10) + 5 in score S on a test based on m questions answered correctly, starting from 10 questions.

explain initial value, factor, and vertical shift.
15 problems Warmup Practice Mixed Review Assessment
Problem 97

Interpret the transformed exponential model T(t)=20+60(0.5)^t in temperature cooling toward room temperature.

Problem 98

Interpret the transformed exponential model P(t)=1000+200(1.1)^t in population above a baseline.

Problem 99

Interpret the transformed exponential model A(t)=5+95(0.8)^t in chemical amount approaching residue level.

Problem 100

Interpret the transformed exponential model M(t)=50+1000(1.03)^t in money in a savings account with a fixed minimum balance.

Problem 101

Interpret the transformed exponential model L(t)=10+50(0.9)^t in pollution level decreasing towards a safe limit.

Problem 102

Interpret the transformed exponential model C(t)=2+8(0.75)^t in drug concentration in the body approaching a steady state.

Problem 103

Interpret the transformed exponential model V(t)=30-20(0.6)^t in object's speed approaching terminal velocity from below.

Problem 104

Interpret the transformed exponential model V(t)=1000+5000(0.85)^t in value of an asset depreciating towards a salvage value.

Problem 105

Interpret the transformed exponential model N(t)=5000-4000(0.95)^t in bacterial population growing towards a carrying capacity.

Problem 106

Interpret the transformed exponential model V(t)=12-12(0.2)^t in voltage in a circuit charging a capacitor.

Problem 107

Interpret the transformed exponential model T(t)=25+75(0.92)^t in temperature of a heated object cooling to ambient.

Open in simulator
Problem 108

Interpret the transformed exponential model R(t)=0.1+10(0.99)^t in amount of a radioactive substance decaying towards a background level.

Problem 109

Interpret the transformed exponential model S(t)=50000+10000(1.05)^t in sales revenue growing above a minimum target.

Problem 110

Interpret the transformed exponential model P(t)=100+20(0.7)^t in pressure in a container approaching an equilibrium pressure.

Problem 111

Interpret the transformed exponential model H(t)=90-40(0.8)^t in health score recovering from illness.

explain differences in rate and starting value.
12 problems Warmup Practice Mixed Review Assessment
Problem 112

Compare the parameter meanings of the linear models A(h)=15h+20 and B(h)=12h+35 in two service plans.

Problem 113

Compare the parameter meanings of the linear models d(t)=60t and e(t)=45t+30 in two trips.

Problem 114

Compare the parameter meanings of the linear models S(w)=25w+100 and T(w)=40w+20 in savings plans.

Problem 115

Compare the parameter meanings of the linear models C1(m)=0.10m+20 and C2(m)=0.05m+30 in two cell phone plans.

Problem 116

Compare the parameter meanings of the linear models H1(d)=0.5d+10 and H2(d)=0.7d+5 in growth of two plants.

Problem 117

Compare the parameter meanings of the linear models F1(d)=-0.05d+15 and F2(d)=-0.04d+12 in fuel remaining in two cars.

Problem 118

Compare the parameter meanings of the linear models W1(t)=-5t+100 and W2(t)=-8t+120 in water draining from two tanks.

Problem 119

Compare the parameter meanings of the linear models E1(s)=0.08s+500 and E2(s)=0.10s+300 in two salespersons' earnings.

Open in simulator
Problem 120

Compare the parameter meanings of the linear models T1(h)=2h+10 and T2(h)=1.5h+15 in temperature changes in two cities.

Problem 121

Compare the parameter meanings of the linear models D1(m)=-100m+1000 and D2(m)=-120m+1200 in debt repayment plans.

Problem 122

Compare the parameter meanings of the linear models R1(d)=50d+150 and R2(d)=40d+200 in two car rental companies.

Problem 123

Compare the parameter meanings of the linear models P1(y)=500y+10000 and P2(y)=400y+12000 in population changes in two towns.

explain differences in initial value and growth factor.
15 problems Warmup Practice Mixed Review Assessment
Problem 124

Compare the parameter meanings of the exponential models A(t)=500(1.04)^t and B(t)=300(1.08)^t in investments.

Open in simulator
Problem 125

Compare the parameter meanings of the exponential models P(t)=1000(0.9)^t and Q(t)=800(0.95)^t in depreciating values.

Problem 126

Compare the parameter meanings of the exponential models M(t)=40(2)^t and N(t)=80(1.5)^t in populations.

Problem 127

Compare the parameter meanings of the exponential models C(t)=200(1.05)^t and D(t)=150(1.10)^t in bacterial colonies.

Problem 128

Compare the parameter meanings of the exponential models X(t)=750(0.85)^t and Y(t)=900(0.92)^t in radioactive decay.

Problem 129

Compare the parameter meanings of the exponential models F(t)=100(1.03)^t and G(t)=100(1.06)^t in investment portfolios.

Problem 130

Compare the parameter meanings of the exponential models K(t)=5000(0.98)^t and L(t)=5000(0.90)^t in car depreciation.

Problem 131

Compare the parameter meanings of the exponential models R(t)=10(3)^t and S(t)=50(1.2)^t in viral spread.

Problem 132

Compare the parameter meanings of the exponential models V(t)=2000(0.5)^t and W(t)=1000(0.75)^t in medication concentration.

Problem 133

Compare the parameter meanings of the exponential models E(t)=10000(1.01)^t and H(t)=8000(1.025)^t in national debt.

Problem 134

Compare the parameter meanings of the exponential models J(t)=1000(0.99)^t and M(t)=1200(0.95)^t in environmental pollutants.

Problem 135

Compare the parameter meanings of the exponential models P(t)=25(2.5)^t and Q(t)=75(1.8)^t in insect populations.

Problem 136

Compare the parameter meanings of the exponential models U(t)=500(0.3)^t and V(t)=300(0.6)^t in chemical reactions.

Problem 137

Compare the parameter meanings of the exponential models A(t)=100(2)^t and B(t)=200(1.5)^t in startup company growth.

Problem 138

Compare the parameter meanings of the exponential models C(t)=800(0.5)^t and D(t)=1200(0.7)^t in drug half-life.

assign units to slope, intercept, initial value, and time exponent.
12 problems Warmup Practice Mixed Review Assessment
Problem 139

Identify the units of the parameters in the model C(h)=18h+25 for cost C in dollars after h hours.

Problem 140

Identify the units of the parameters in the model P(t)=600(1.03)^t for population P after t years.

Problem 141

Identify the units of the parameters in the model d(m)=0.4m+2 for distance d in miles after m minutes.

Problem 142

Identify the units of the parameters in the model T(s)=3s+10 for temperature T in degrees Celsius after s seconds.

Problem 143

Identify the units of the parameters in the model A(w)=5000(0.95)^w for amount A in grams after w weeks.

Problem 144

Identify the units of the parameters in the model V(t)=-2.5t+100 for volume V in liters after t minutes.

Problem 145

Identify the units of the parameters in the model B(h)=100(1.12)^h for bacteria count B after h hours.

Problem 146

Identify the units of the parameters in the model F(x)=0.75x+5 for fuel F in gallons after x miles.

Problem 147

Identify the units of the parameters in the model R(d)=200(0.8)^d for remaining R in pages after d days.

Problem 148

Identify the units of the parameters in the model W(m)=0.05m+15 for weight W in pounds after m months.

Problem 149

Identify the units of the parameters in the model S(y)=15000(1.04)^y for salary S in dollars after y years.

Open in simulator
Problem 150

Identify the units of the parameters in the model H(k)=-5k+50 for height H in feet after k kilometers.

detect rate/intercept/factor confusion.
12 problems Warmup Practice Mixed Review Assessment
Problem 151

Correct the parameter interpretation error: 50 means the cost increases by 50 dollars per hour for C(h)=20h+50.

Problem 152

Correct the parameter interpretation error: 1.06 means add 1.06 people each year for P(t)=300(1.06)^t.

Problem 153

Correct the parameter interpretation error: 0.8 means the starting value is 0.8 for V(t)=100(0.8)^t.

Problem 154

Correct the parameter interpretation error: 3 is the initial value for y = 3x + 10.

Open in simulator
Problem 155

Correct the parameter interpretation error: 2 is the starting amount for f(x) = 50 - 2x.

Problem 156

Correct the parameter interpretation error: 1.03 means 1.03 units are added each time period for A(t) = 500 * (1.03)^t.

Problem 157

Correct the parameter interpretation error: 0.95 is the starting population for P(x) = 1000 * (0.95)^x.

Problem 158

Correct the parameter interpretation error: 15 is the cost per mile for Cost = 15 + 5 * miles.

Problem 159

Correct the parameter interpretation error: 1.15 means the value increases by 1.15 units for Value = 200 * (1.15)^t.

Problem 160

Correct the parameter interpretation error: 75 is how much the temperature decreases per minute for T(m) = 75 - 0.5m.

Problem 161

Correct the parameter interpretation error: 0.7 means the amount decreases by 70% each hour for D(h) = 80 * (0.7)^h.

Problem 162

Correct the parameter interpretation error: 100 is the rate of increase for R(x) = 12x + 100.