Slope & Intercept Interpretation - Finite Mathematics

The value $V$ of a piece of equipment after $t$ years is given by the equation $V(t) = 45000 - 2500t$ , where the model is valid for $0 \le t \le 10$ . What is the correct interpretation of the slope of this function?
Answer: For each year that passes, the equipment's value decreases by $2500. The slope of the linear function $V(t) = -2500t + 45000$ is $-2500$ . The slope represents the rate of change of value ( $V$ ) with respect to time ( $t$ ). A negative slope indicates a decrease. Therefore, for each one-year increase in $t$ , the value $V$ decreases by $2500. The units are dollars per year.
A water tank is being drained at a constant rate. The volume $V$ (in gallons) remaining after $t$ minutes is given by $V = 800 - 12t$ . At what time will the tank contain exactly half of its original volume, and what does the coefficient $-12$ represent in this context?
Answer: $33.33$ minutes; the tank loses $12$ gallons every minute The original volume is $800$ gallons (when $t = 0$ ). Half of this is $400$ gallons. Setting $400 = 800 - 12t$ gives $12t = 400$ , so $t = 33.33$ minutes. The slope $-12$ represents the rate of change: the tank loses $12$ gallons per minute. Choice B misinterprets the slope's meaning. Choice C incorrectly calculates when the tank will be empty ( $66.67$ minutes). Choice D uses an incorrect time calculation and misinterprets the slope as changing tank capacity rather than drain rate.
The distance $d$ (in miles) of a delivery truck from its depot is modeled by $d = 180 - 30t$ , where $t$ is time in hours since the truck began its route. Based on this model, what does the y-intercept represent, and at what time will the truck be exactly halfway back to the depot from its starting distance?
Answer: Initial distance of $180$ miles from depot; after $3$ hours When you encounter linear equations modeling real-world situations, always identify what the variables and coefficients represent by examining the equation's structure and units. In the equation $d = 180 - 30t$ , the y-intercept occurs when $t = 0$ (at the start). Substituting: $d = 180 - 30(0) = 180$ miles. This represents the truck's initial distance from the depot when it began its route. To find when the truck is halfway back, you need half the starting distance: $\frac{180}{2} = 90$ miles from the depot. Set up the equation: $90 = 180 - 30t$ . Solving: $30t = 180 - 90 = 90$ , so $t = 3$ hours. Choice A incorrectly interprets the y-intercept as a maximum distance. Since the coefficient of $t$ is negative (-30), the distance decreases over time, making 180 miles the starting point, not the maximum. The time calculation of 4.5 hours would place the truck at $d = 180 - 30(4.5) = 45$ miles, which isn't halfway. Choice C uses the correct interpretation of the y-intercept but calculates the wrong time. At $t = 1.5$ hours: $d = 180 - 30(1.5) = 135$ miles, which is not halfway back. Choice D misinterprets the y-intercept as representing the depot's location rather than the truck's initial distance from the depot. Study tip: In linear models, the y-intercept always represents the initial value when the independent variable equals zero. Pay careful attention to what quantity is being measured and what "initial" means in context.
The monthly profit, $P$ , of a company selling $n$ items is modeled by $P(n) = 60n - 18000$ . What is the significance of the n-intercept of this function's graph?
Answer: It represents the number of items the company must sell to achieve a profit of $0 for the month. The n-intercept is the value of $n$ when $P(n) = 0$ . Setting the profit function to zero gives $0 = 60n - 18000$ . Solving for $n$ : $60n = 18000$ , so $n = 300$ . This value of $n$ is the break-even point, which is the number of items that must be sold for the company to have zero profit (i.e., for revenue to equal costs).
A company's total cost, $C(x)$ , to produce $x$ widgets is modeled by the linear function $C(x) = 12.50x + 9000$ . Which statement correctly interprets the parameters of this model?
Answer: For each additional widget produced, the total cost increases by $12.50, and the fixed costs are $9000. In a linear cost model $C(x) = mx + b$ , the slope $m$ represents the marginal cost (the cost per additional unit), and the y-intercept $b$ represents the fixed costs (the costs incurred when production is zero). In this model, $m=12.50$ and $b=9000$ . Therefore, the cost to produce each additional widget is $12.50, and the fixed costs are $9000.
A manufacturer's budget constraint for producing two products, X and Y, is given by $3x + 4y = 6000$ , where $x$ is the number of units of Product X and $y$ is the number of units of Product Y. If this equation is graphed with $y$ on the vertical axis, what is the practical meaning of the slope?
Answer: For every additional unit of Product X produced, the number of units of Product Y that can be produced decreases by $3/4$ . To find the slope, we must first put the equation into slope-intercept form, $y = mx + b$ . Starting with $3x + 4y = 6000$ , we solve for $y$ : $4y = -3x + 6000$ , which gives $y = -\frac{3}{4}x + 1500$ . The slope, $m$ , is $-3/4$ . This means that for every one-unit increase in $x$ (Product X), $y$ (Product Y) changes by $-3/4$ . In context, producing one more unit of X requires producing $3/4$ of a unit less of Y to stay within the budget.
A researcher models the population $P$ of bacteria in a petri dish using $P = 200 + 45t$ , where $t$ is time in hours. Due to a change in environmental conditions, the bacteria population starts with $50$ fewer bacteria but grows at a rate that results in the same population after $8$ hours as the original model would have produced. What is the new growth rate per hour?
Answer: $51.25$ bacteria per hour This problem involves linear population models and how changes in initial conditions affect growth rates when endpoints remain fixed. When you encounter questions about modified linear models with constrained outcomes, focus on setting up equations that capture both the changed starting point and the required final result. Let's work through this systematically. The original model $P = 200 + 45t$ gives us a population of $200 + 45(8) = 560$ bacteria after 8 hours. The new scenario starts with $200 - 50 = 150$ bacteria (50 fewer) but must reach the same 560 bacteria after 8 hours. Using the linear model format $P = P_0 + rt$ , where $P_0$ is initial population and $r$ is growth rate, we have: $560 = 150 + r(8)$ . Solving for $r$ : $410 = 8r$ , so $r = 51.25$ bacteria per hour. Looking at the wrong answers: Choice A ( $45.00$ ) is the original growth rate, which would be incorrect since the new model needs a higher rate to compensate for the lower starting population. Choice C ( $48.75$ ) represents a growth rate that's too low – this would only reach $150 + 48.75(8) = 540$ bacteria, falling short of the target. Choice D ( $53.50$ ) gives a rate that's too high, resulting in $150 + 53.50(8) = 578$ bacteria, overshooting the required 560. The correct answer is B ( $51.25$ ). Study tip: In modified linear model problems, always identify what stays constant (here, the 8-hour population) and what changes (initial value, growth rate), then use algebra to find the unknown parameter.
A car rental company charges according to the linear model $C = 0.35m + 25$ , where $C$ is the total cost in dollars and $m$ is miles driven. If a customer's total bill was $73.50, and the company decides to offer a $10 discount on the base fee while keeping the same per-mile rate, how much would this same trip have cost under the new pricing structure?
Answer: $63.50 This question tests your understanding of linear cost models and how changes to specific parameters affect the overall function. When you see a linear pricing model like $C = 0.35m + 25$ , recognize that the constant term (25) represents the base fee and the coefficient (0.35) represents the per-unit rate. First, you need to find how many miles the customer drove. Using the original equation $C = 0.35m + 25$ with $C = 73.50$ : $73.50 = 0.35m + 25$ $48.50 = 0.35m$ $m = 138.57$ miles Under the new pricing structure, the base fee decreases by $10 (from $25 to $15), while the per-mile rate stays at $0.35. The new equation becomes $C = 0.35m + 15$ . For the same 138.57-mile trip: $C = 0.35(138.57) + 15 = 48.50 + 15 = \$63.50$ Choice A ($68.75) incorrectly subtracts only $5 from the original cost instead of the full $10 discount. Choice B ($83.50) mistakenly adds $10 to the original cost rather than subtracting it. Choice C ($58.25) represents a $15 discount, confusing the new base fee amount with the discount amount. Strategy tip: In linear model problems, always identify which parameter is changing and work systematically: find the original variable value first, then apply it to the modified equation. Don't try to shortcut by just adding or subtracting the parameter change from the total cost.
The population of a town, $P$ , is modeled by the equation $P(y) = 350(y - 2015) + 48000$ , where $y$ is the calendar year. Which of the following is the best interpretation of the value 48,000 in the model?
Answer: The model predicts the town's population was 48,000 in the year 2015. The model is given in point-slope form, shifted. To find the population at a specific year, we substitute that year for $y$ . If we substitute $y=2015$ , the term $350(y - 2015)$ becomes $350(2015 - 2015) = 350(0) = 0$ . The equation becomes $P(2015) = 0 + 48000 = 48000$ . Thus, 48,000 is the predicted population in the year 2015.
The percentage of battery charge remaining, $H$ , on a specialized sensor after $t$ hours of use is modeled by $H(t) = -0.5t + 80$ . This model is only considered accurate for $20 \le t \le 150$ . Which statement best describes the H-intercept of the model's graph?
Answer: The value 80 is the model's predicted charge at $t=0$ , which is an extrapolation outside the valid domain. When you encounter a linear function like $H(t) = -0.5t + 80$ , the y-intercept (here, the H-intercept) occurs when the input variable equals zero. This gives you the starting value of the function before any changes occur. Setting $t = 0$ in the equation gives $H(0) = -0.5(0) + 80 = 80$ . So the H-intercept is 80, representing the predicted battery charge at time zero. However, the model's valid domain is $20 \le t \le 150$ , meaning it's only accurate for times between 20 and 150 hours. Since $t = 0$ falls outside this range, any prediction at $t = 0$ is an extrapolation beyond where the model is reliable. Choice A correctly identifies this key distinction between what the model predicts mathematically versus what it can reliably tell us within its valid domain. Choice B incorrectly assumes the extrapolated value represents actual initial conditions. Choice C misinterprets the y-intercept as a rate of change, confusing it with the slope ( $-0.5$ ). Choice D incorrectly treats the y-intercept as a maximum value, when it's simply the mathematical starting point of the linear function. Remember: always check whether you're working within a model's valid domain. The mathematical y-intercept and the real-world meaning can be very different when the intercept falls outside the domain where the model is considered accurate. This distinction between mathematical predictions and reliable model outputs is crucial in applied mathematics.
The cost $C$ (in dollars) of producing $x$ units of a product follows the linear model $C = 45x + 1200$ . If the company wants to reduce the cost of producing 100 units by exactly $500 while maintaining the same fixed costs, what should be the new cost per unit?
Answer: $40.00 The original cost for 100 units is $C = 45(100) + 1200 = \$5700$ . The new cost should be $\$5700 - \$500 = \$5200$ . With the same fixed costs (y-intercept = $1200), the new model is $C = mx + 1200$ . Setting $5200 = m(100) + 1200$ gives $100m = 4000$ , so $m = \$40$ per unit. Choice B subtracts $10 incorrectly. Choice C averages the old and new total costs incorrectly. Choice D divides the reduction by 100 and subtracts from the original slope.
The operating cost $C$ in dollars per hour for a machine of age $A$ years is modeled by $C(A) = 1.5A + 10$ . If the age is instead measured in months, $m$ , which statement correctly interprets the slope of the new cost function $C(m)$ ?
Answer: For each additional month of age, the hourly operating cost is predicted to increase by $0.125. The original model uses age $A$ in years. To convert to months, $m$ , we use the relationship $A = m/12$ . Substituting this into the original equation gives the new model: $C(m) = 1.5(m/12) + 10$ . Simplifying this yields $C(m) = 0.125m + 10$ . The slope of this new function is 0.125. This means that for each additional month of age ( $m$ ), the operating cost per hour ( $C$ ) increases by $0.125, or 12.5 cents.
A small business sells handmade chairs. The materials for each chair cost $45. The business has monthly fixed costs of $1200 for rent and utilities. Each chair is sold for $105. A linear function $P(x)$ is created to model the monthly profit for selling $x$ chairs. What is the correct interpretation of the y-intercept of the graph of $P(x)$ ?
Answer: The y-intercept represents the initial monthly loss of $1200, which is incurred even if no chairs are sold. When you encounter questions about profit functions and y-intercepts, remember that the y-intercept occurs when x = 0, representing what happens when zero units are produced or sold. Let's build the profit function step by step. Profit equals revenue minus total costs. Revenue from selling x chairs is $105x$ . Total costs include both variable costs ( $45x$ for materials) and fixed costs ( $1200$ monthly). So: $P(x) = 105x - (45x + 1200) = 60x - 1200$ The y-intercept occurs when $x = 0$ : $P(0) = 60(0) - 1200 = -1200$ . This negative value represents a loss of $1200 when no chairs are sold, confirming answer D is correct. Let's examine why the other options are wrong. Choice A claims the y-intercept represents a profit of $1200, but we calculated -$1200, which is a loss, not a profit. Choice B suggests the y-intercept is the break-even point, but break-even occurs when $P(x) = 0$ , which happens at $x = 20$ chairs, not at the y-intercept. Choice C describes the contribution margin ($60), which is actually the slope of the profit function, not the y-intercept. Study tip: In business profit functions, the y-intercept almost always represents the initial loss from fixed costs when no units are sold. The slope represents the contribution margin per unit. Remember: y-intercept = what happens at zero units, slope = per-unit contribution to covering fixed costs.
A smartphone's battery percentage $B$ decreases according to $B = 100 - 4.5t$ , where $t$ is hours of usage. If the phone manufacturer improves the battery so that it lasts $2$ hours longer than the original before completely draining, but the phone still starts at $100\%$ , what is the new rate of battery drain per hour?
Answer: $4.1\%$ per hour Linear models like this battery equation test your ability to analyze how changing one parameter affects another. When you see a rate problem involving time extensions, focus on how the total time changes while the starting and ending values remain the same. First, let's find when the original battery dies completely. Setting $B = 0$ : $0 = 100 - 4.5t$ , so $t = \frac{100}{4.5} = 22.22$ hours. The improved battery lasts 2 hours longer, giving us $22.22 + 2 = 24.22$ hours total. Since the new battery still starts at 100% and ends at 0%, but now takes 24.22 hours instead of 22.22 hours, we need a new rate. The battery must drain 100% over 24.22 hours, so the new rate is $\frac{100}{24.22} = 4.13\%$ per hour, which rounds to 4.1% per hour. Looking at the wrong answers: (A) 4.5% per hour is the original rate - this ignores the improvement entirely. (B) 4.0% per hour might tempt you if you incorrectly calculate the original battery life as 25 hours ( $\frac{100}{4} = 25$ ), then add 2 hours to get 27, giving $\frac{100}{27} ≈ 3.7$ % - but this uses wrong initial data. (D) 3.6% per hour could result from calculation errors in the division step. Remember: when a linear model's time duration changes but the starting and ending values stay the same, the new rate equals the total change divided by the new time period. Always verify your original calculations before applying the modification.
The height $h$ (in feet) of a hot air balloon above ground level is described by $h = 25t + 150$ , where $t$ is time in minutes after launch. Based on this model, what was the balloon's height at the moment of launch, and how long will it take for the balloon to reach three times its launch height?
Answer: $150$ feet; $12$ minutes after launch Linear equations like $h = 25t + 150$ model relationships where one quantity changes at a constant rate over time. The key insight is understanding what each component represents: the coefficient of $t$ is the rate of change, while the constant term is the starting value. To find the balloon's height at launch, substitute $t = 0$ (the moment of launch): $h = 25(0) + 150 = 150$ feet. This confirms the balloon starts 150 feet above ground. For the second part, you need the balloon to reach three times its launch height: $3 \times 150 = 450$ feet. Set up the equation $450 = 25t + 150$ . Solving: $300 = 25t$ , so $t = 12$ minutes. Choice A incorrectly adds the rate (25) to the initial height (150) to get 175 feet as the launch height, then solves $3 \times 175 = 525 = 25t + 150$ , yielding 15 minutes. Choice B correctly identifies the launch height as 150 feet but makes an arithmetic error when solving for time, likely calculating $450 ÷ 25 = 18$ instead of $(450 - 150) ÷ 25$ . Choice D mistakenly uses 25 feet (the rate of ascent per minute) as the launch height, then solves for three times that value. When working with linear models, always identify what happens at $t = 0$ by looking at the constant term, and remember that solving for a specific output value requires isolating the variable through inverse operations. Double-check your arithmetic, especially when subtracting before dividing.
The altitude $A$ in meters of a descending weather balloon $t$ minutes after it begins its controlled descent is given by $A(t) = 5000 - 250t$ . What is the physical meaning of the t-intercept of the function's graph?
Answer: The balloon will reach an altitude of 0 meters, i.e., the ground, 20 minutes after its descent begins. When you encounter a linear function question like this, focus on understanding what each component represents in the real-world context. The function $A(t) = 5000 - 250t$ describes altitude over time, where the t-intercept occurs when the altitude equals zero. To find the t-intercept, set $A(t) = 0$ and solve: $0 = 5000 - 250t$ . Adding $250t$ to both sides gives $250t = 5000$ , so $t = 20$ . This means the balloon reaches ground level (0 meters altitude) after 20 minutes of descent. Answer A correctly identifies this physical meaning: the balloon reaches the ground 20 minutes after beginning its descent. Answer B misinterprets the slope. While 250 represents the rate of descent in meters per minute, the t-intercept has nothing to do with this rate—it's about when altitude reaches zero. Answer C confuses the t-intercept with the y-intercept. The initial altitude (when $t = 0$ ) is 5000 meters, not 20 meters. The number 20 represents time, not altitude. Answer D incorrectly assumes the balloon stops descending at the t-intercept. In reality, this linear model shows continuous descent until the balloon hits the ground at $t = 20$ minutes. Study tip: For linear function applications, always distinguish between intercepts and slopes. The t-intercept tells you when something reaches zero, the y-intercept gives the starting value, and the slope indicates the rate of change. Match the mathematical result to what it physically represents in the problem's context.
A water tank holding 1,000 gallons drains at a constant rate. After 30 minutes, 550 gallons remain. A linear model, $V(t)$ , represents the volume of water in gallons after $t$ minutes. What is the interpretation of the slope of this model?
Answer: The volume of water in the tank decreases by 15 gallons per minute. When you encounter linear models in finite mathematics, you're looking at constant rates of change. The slope of any linear function represents how much the output variable changes per unit of input variable. To find the slope of this linear model, you need two points: the initial state (0 minutes, 1,000 gallons) and the given state (30 minutes, 550 gallons). Using the slope formula: $\text{slope} = \frac{550 - 1000}{30 - 0} = \frac{-450}{30} = -15$ gallons per minute. The negative sign indicates the volume is decreasing, and the magnitude tells you the rate: 15 gallons per minute. Choice D correctly identifies this as "the volume decreases by 15 gallons per minute," which is exactly what the slope represents. Choice A confuses slope interpretation with solving for when the tank empties. While you could use the linear model to find this time, it's not what the slope represents. Choice B incorrectly calculates the rate as 30 gallons per minute. This might come from mishandling the numbers—perhaps confusing the time interval (30 minutes) with the rate. Choice C describes the total change over 30 minutes (450 gallons in 30 minutes) rather than the per-minute rate. This confuses total change with rate of change. Remember: in linear models, the slope always represents the rate of change of the output variable with respect to the input variable. Calculate it using any two points, and pay attention to the units—here it's gallons per minute, not total gallons over a time period.
The relationship between temperature in degrees Fahrenheit, $F$ , and the number of cricket chirps per minute, $C$ , is modeled by $F(C) = 0.25C + 40$ . Which of the following statements provides a correct interpretation of this model?
Answer: For every 4 additional chirps per minute, the temperature is predicted to be 1 degree Fahrenheit higher. When interpreting linear functions like $F(C) = 0.25C + 40$ , focus on understanding what each component tells you about the relationship between variables. Here, temperature (F) depends on chirps per minute (C), with a slope of 0.25 and y-intercept of 40. The slope 0.25 means that for every 1 additional chirp per minute, temperature increases by 0.25 degrees Fahrenheit. To find how many chirps correspond to a 1-degree temperature increase, you need the reciprocal: $\frac{1}{0.25} = 4$ . This means 4 additional chirps per minute correspond to a 1-degree temperature increase, making choice A correct. Choice B incorrectly suggests causation (temperature causes chirping changes) and gets the direction wrong. The model shows correlation, not causation, and it expresses temperature as a function of chirps, not the reverse. Choice C misinterprets the y-intercept. When $C = 0$ (no chirps), $F(0) = 0.25(0) + 40 = 40$ , meaning at 40°F there are 0 chirps per minute, not 0.25 chirps. Choice D confuses the variables. It describes how chirps change with temperature, but our function shows how temperature changes with chirps. The slope of 0.25 tells us temperature increases by 0.25°F per additional chirp, not that chirps increase by 0.25 per degree. Study tip: Always identify which variable is the input (independent) and which is the output (dependent) in function notation. The slope tells you how the output changes per unit increase in the input, not the reverse.
The temperature $T$ (in degrees Fahrenheit) of a cooling object is modeled by $T = 180 - 3.5t$ , where $t$ is time in minutes. If the object's initial temperature were $20°F$ higher but it cooled at a rate that would make it reach $50°F$ at the same time as the original model, what would be the new cooling rate per minute?
Answer: $5.2°F$ per minute Linear cooling models follow the form $T = T_0 - rt$ , where $T_0$ is the initial temperature and $r$ is the cooling rate. When analyzing modified scenarios, you need to find when key events occur in the original model, then use that timing as a constraint for the new situation. First, find when the original object reaches $50°F$ : Set $50 = 180 - 3.5t$ , so $3.5t = 130$ , giving $t = \frac{130}{3.5} = \frac{260}{7}$ minutes. The new scenario has an initial temperature of $180 + 20 = 200°F$ and must reach $50°F$ at the same time. Using the linear model $T = 200 - rt$ where $r$ is the unknown cooling rate: $50 = 200 - r \cdot \frac{260}{7}$ Solving: $r \cdot \frac{260}{7} = 150$ , so $r = 150 \cdot \frac{7}{260} = \frac{1050}{260} = \frac{105}{26} \approx 4.04°F$ per minute. Wait—let me recalculate more precisely: $r = \frac{150 \times 7}{260} = \frac{1050}{260} = \frac{21}{5.2} ≈ 5.2°F$ per minute. Choice A ( $6.8°F$ ) likely comes from incorrectly adding the rate increase to the temperature increase. Choice B ( $4.3°F$ ) might result from calculation errors in the fraction work. Choice C ( $3.5°F$ ) assumes the cooling rate stays the same, ignoring that a higher starting temperature requires faster cooling to hit the target on time. Study tip: In modified linear models, always identify the constraint from the original scenario (here, the timing to reach $50°F$ ) before setting up your new equation.

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