Math 3 Quiz: Domain And Discontinuities
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Domain And DiscontinuitiesQuestion 1 of 20

A water treatment plant processes water through a filtration system. The efficiency of the system is modeled by the function E(t) = (500t - 1000)/(t² - 4t + 4), where E(t) represents the percentage efficiency and t represents time in hours since startup.

The plant manager notices that the efficiency function becomes undefined at a specific time. What is the practical interpretation of this discontinuity in the context of the water treatment process?

At t = 2 hours, the system reaches maximum efficiency and temporarily shuts down for maintenance
At t = 2 hours, the filtration rate equals the input rate, causing the efficiency calculation to be undefined
At t = 4 hours, the system experiences a complete failure due to filter saturation
At t = 1 hour, the system transitions from startup mode to normal operating mode
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Math 3 Quiz

Math 3 Quiz: Domain And Discontinuities

Practice Domain And Discontinuities in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Domain And Discontinuities, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A water treatment plant processes water through a filtration system. The efficiency of the system is modeled by the function E(t) = (500t - 1000)/(t² - 4t + 4), where E(t) represents the percentage efficiency and t represents time in hours since startup.

The plant manager notices that the efficiency function becomes undefined at a specific time. What is the practical interpretation of this discontinuity in the context of the water treatment process?

  1. At t = 2 hours, the system reaches maximum efficiency and temporarily shuts down for maintenance
  2. At t = 2 hours, the filtration rate equals the input rate, causing the efficiency calculation to be undefined (correct answer)
  3. At t = 4 hours, the system experiences a complete failure due to filter saturation
  4. At t = 1 hour, the system transitions from startup mode to normal operating mode
Explanation: The function E(t) = (500t - 1000)/(t² - 4t + 4) has a denominator that factors as (t - 2)², which equals zero when t = 2. This creates a discontinuity at t = 2 hours. In the context of efficiency calculations, this suggests that at t = 2, the denominator of the efficiency ratio becomes zero, which would occur when the filtration rate equals the input rate, making the efficiency percentage calculation undefined.

Question 2

An environmental scientist models the oxygen level in a lake using O(d) = (d² - 25)/(d² + 3d - 10), where O(d) represents oxygen concentration in ppm and d represents depth in meters below the surface.

The scientist needs to identify depths where the oxygen model becomes invalid. At which depth(s) does the model predict problematic conditions, and what do these represent physically?

  1. At d = 2 meters, a vertical asymptote suggests oxygen levels become infinitely high due to algae concentration
  2. At d = -5 meters, a domain restriction occurs, but this is physically meaningless since depth cannot be negative
  3. At d = 2 meters, a vertical asymptote indicates the model breaks down, possibly due to a thermocline layer (correct answer)
  4. At d = 5 meters, a removable discontinuity suggests a temporary oxygen depletion that can be corrected
Explanation: The denominator d² + 3d - 10 = (d+5)(d-2) equals zero at d = -5 and d = 2. Since d = -5 (negative depth) is physically meaningless, only d = 2 is relevant. The numerator d² - 25 = (d-5)(d+5) equals 25 - 25 = 0 at d = 5, not at d = 2. At d = 2, the numerator is 4 - 25 = -21 ≠ 0, so there's a vertical asymptote, suggesting the model breaks down at 2 meters depth.

Question 3

An economics model predicts that the cost per unit C(x) for manufacturing x thousand items is given by C(x) = (2x² + 8x)/(x² - x - 6). The company's production manager needs to understand where this cost function is undefined.

If the company plans to produce anywhere from 1 to 5 thousand items, which production level(s) should be avoided according to the domain restrictions of the cost function?

  1. Production should avoid exactly 3 thousand items due to a vertical asymptote in the cost function (correct answer)
  2. Production should avoid both 2 and 3 thousand items due to domain restrictions in the cost function
  3. Production should avoid exactly 2 thousand items due to a removable discontinuity in the cost function
  4. Production levels between 1 and 5 thousand items are all acceptable since no domain restrictions exist in this range
Explanation: To find domain restrictions, set the denominator equal to zero: x² - x - 6 = 0. Factoring gives (x - 3)(x + 2) = 0, so x = 3 or x = -2. Within the production range of 1 to 5 thousand items, only x = 3 is relevant. Since the numerator 2x² + 8x = 2x(x + 4) doesn't equal zero at x = 3, this creates a vertical asymptote, meaning the cost becomes infinitely large at 3 thousand items.

Question 4

A marketing team models the effectiveness E(t) of an advertising campaign using E(t) = (t² - t - 2)/(t² + t - 6), where t represents weeks since launch and E(t) represents effectiveness as a decimal.

The team discovers that their effectiveness model produces undefined results at certain time periods. Which weeks require special consideration, and what do these discontinuities suggest about the campaign?

  1. Week 2 has a removable discontinuity suggesting temporary data collection issues, while week -3 is irrelevant for future planning (correct answer)
  2. Week 2 has a vertical asymptote indicating maximum effectiveness, while week -3 represents pre-launch baseline conditions
  3. Weeks 2 and -3 both have vertical asymptotes, but only week 2 is relevant for ongoing campaign management decisions
  4. Week -1 has a removable discontinuity in the numerator, while weeks 2 and -3 have domain restrictions from the denominator
Explanation: Factor both polynomials: numerator t² - t - 2 = (t - 2)(t + 1), denominator t² + t - 6 = (t + 3)(t - 2). So E(t) = (t - 2)(t + 1)/((t + 3)(t - 2)). Domain restrictions occur where the denominator equals zero: t = -3 and t = 2. At t = 2, both numerator and denominator have factor (t - 2), creating a removable discontinuity. At t = -3, only the denominator is zero, creating a vertical asymptote. Since t = -3 represents 3 weeks before launch, it's not relevant for campaign management.

Question 5

The function f(x) = (x² - 9)/(x² + 2x - 15) represents a rational function. A student claims that the domain restrictions occur at x = -5 and x = 3, and that both create vertical asymptotes. Which statement best evaluates this claim?

  1. The claim is correct; both x = -5 and x = 3 are domain restrictions that create vertical asymptotes
  2. The claim is incorrect; both x = -5 and x = 3 create removable discontinuities, not vertical asymptotes
  3. The claim is partially correct; x = 3 creates a vertical asymptote, but x = -5 creates a removable discontinuity
  4. The claim is partially correct; x = -5 creates a vertical asymptote, but x = 3 creates a removable discontinuity (correct answer)
Explanation: When analyzing rational functions for domain restrictions and discontinuities, you need to factor both the numerator and denominator, then determine what happens at each zero of the denominator. Let's factor this function: f(x)=x2−9x2+2x−15=(x−3)(x+3)(x+5)(x−3)f(x) = \frac{x^2 - 9}{x^2 + 2x - 15} = \frac{(x-3)(x+3)}{(x+5)(x-3)} The domain restrictions occur where the denominator equals zero: at x = -5 and x = 3. However, these create different types of discontinuities. At x = -5: Only the denominator has a factor of (x+5), so this creates a vertical asymptote where the function approaches ±∞. At x = 3: Both numerator and denominator have the factor (x-3), which cancels out. This creates a removable discontinuity (a "hole") rather than a vertical asymptote. The simplified function becomes f(x)=x+3x+5f(x) = \frac{x+3}{x+5} for x ≠ 3. Answer choice A incorrectly claims both points create vertical asymptotes. Answer choice B incorrectly states both create removable discontinuities. Answer choice C reverses which point creates which type of discontinuity. Answer choice D correctly identifies that x = -5 creates a vertical asymptote while x = 3 creates a removable discontinuity. Study tip: Always factor rational functions completely and look for common factors between numerator and denominator. Common factors create holes (removable discontinuities), while zeros that appear only in the denominator create vertical asymptotes.

Question 6

An agricultural scientist studies crop yield efficiency using the model Y(f) = (f² - 9)/(f² - 2f - 3), where Y(f) represents yield ratio and f represents fertilizer concentration in kg per hectare.

The scientist needs to determine fertilizer concentrations that make the yield model invalid. What concentration should be avoided, and what does this restriction imply about optimal fertilization strategy?

  1. Avoid f = 3 kg/ha due to a vertical asymptote; this suggests over-fertilization leads to infinite yield ratios
  2. Avoid f = 3 kg/ha due to a removable discontinuity; the yield can be calculated using limits at this concentration (correct answer)
  3. Avoid f = -1 kg/ha due to a vertical asymptote, but negative fertilizer concentrations are physically meaningless
  4. Avoid both f = 3 and f = -1 kg/ha; both create vertical asymptotes indicating model breakdown at these concentrations
Explanation: Factor the polynomials: numerator f² - 9 = (f - 3)(f + 3), denominator f² - 2f - 3 = (f - 3)(f + 1). Domain restrictions occur at f = 3 and f = -1. At f = 3, both numerator and denominator have factor (f - 3), creating a removable discontinuity. The limit as f approaches 3 can be calculated by canceling the common factor: Y(f) = (f + 3)/(f + 1), which gives (3 + 3)/(3 + 1) = 6/4 = 1.5 at f = 3. At f = -1, only the denominator is zero, creating a vertical asymptote, but negative fertilizer concentration is not physically meaningful.

Question 7

The rational function h(x) = (x² - 2x - 8)/(x³ - 2x² - 8x) has multiple domain restrictions. Which of the following correctly identifies all domain restrictions and their corresponding types of discontinuities?

  1. Domain restrictions at x = 0, 4, -2; all three create vertical asymptotes due to zero denominators
  2. Domain restrictions at x = 0, 4, -2; x = 0 and x = -2 create vertical asymptotes, x = 4 creates a removable discontinuity (correct answer)
  3. Domain restrictions at x = 0, 4, -2; x = 0 creates a vertical asymptote, x = 4 and x = -2 create removable discontinuities
  4. Domain restrictions at x = 0, 4, -2; x = 4 creates a vertical asymptote, x = 0 and x = -2 create removable discontinuities
Explanation: Factor both polynomials: numerator x² - 2x - 8 = (x-4)(x+2), denominator x³ - 2x² - 8x = x(x² - 2x - 8) = x(x-4)(x+2). So h(x) = (x-4)(x+2)/(x(x-4)(x+2)). Domain restrictions occur at x = 0, 4, -2. At x = 4 and x = -2, common factors in numerator and denominator create removable discontinuities. At x = 0, only the denominator has factor x, creating a vertical asymptote.

Question 8

A chemical engineer models the concentration C(r) of a reactant using C(r) = (r³ + 8)/(r² + 2r), where r represents reaction time in minutes and C(r) represents concentration in mol/L.

The engineer observes unusual behavior in the concentration model at specific reaction times. Which time point requires the most careful analysis for process safety?

  1. At r = 0 minutes, the model predicts infinite concentration, requiring immediate safety protocols before reaction start (correct answer)
  2. At r = -2 minutes, the model shows a removable discontinuity, but negative time values are not physically meaningful
  3. At r = 0 minutes, there's a removable discontinuity that can be resolved by calculating the limit as r approaches zero
  4. At r = -2 minutes, there's a vertical asymptote indicating dangerous concentration levels during pre-reaction preparation
Explanation: The denominator r² + 2r = r(r + 2) equals zero at r = 0 and r = -2. The numerator r³ + 8 = (r + 2)(r² - 2r + 4). At r = 0, the numerator equals 0³ + 8 = 8 ≠ 0, while the denominator equals 0. This creates a vertical asymptote at r = 0, meaning concentration approaches infinity as time approaches zero. At r = -2, both numerator and denominator have factor (r + 2), creating a removable discontinuity. Since r = 0 represents the start of the reaction and shows infinite concentration, this requires immediate safety consideration.

Question 9

A manufacturer models the cost per unit (in dollars) for producing widgets as C(x)=500x+12000x−50C(x) = \frac{500x + 12000}{x - 50}, where xx is the number of units produced. The company's quality control department notes that production levels consistently avoid certain values. Which statement best explains the relationship between the mathematical domain restriction and the physical meaning in this context?

  1. Production cannot equal 50 units because the cost per unit becomes undefined, representing an impossible manufacturing scenario where fixed costs cannot be distributed. (correct answer)
  2. Production cannot exceed 50 units because the cost function becomes negative, which would mean the company pays customers to take the widgets.
  3. Production cannot equal 50 units because this creates a removable discontinuity where costs can be calculated using limits but not direct substitution.
  4. Production cannot be less than 50 units because the denominator becomes negative, representing a scenario where the company loses money on each widget.
Explanation: At x = 50, the denominator equals zero, creating a vertical asymptote and making the cost per unit undefined. This represents a mathematical impossibility in the real-world context - you cannot calculate a meaningful cost per unit at this production level. Choice B is wrong because the function doesn't become negative for x > 50. Choice C is incorrect because this is a non-removable discontinuity (vertical asymptote), not removable. Choice D is wrong because negative denominators don't necessarily mean losing money, and the function is defined for x < 50.

Question 10

An environmental scientist models the concentration of a pollutant in a lake as P(t)=100tt2−25P(t) = \frac{100t}{t^2 - 25}, where P(t)P(t) is the concentration in parts per million and tt is time in days after monitoring begins.

The scientist observes that the model becomes unrealistic at certain time values. Which interpretation best explains why the domain restriction at t=5t = 5 is significant in this environmental context?

  1. At day 5, the concentration approaches infinity, suggesting an environmental catastrophe where pollutant levels spike beyond measurable limits due to system breakdown. (correct answer)
  2. At day 5, the model predicts negative concentration, which is physically impossible since pollutant concentration cannot be less than zero in real water systems.
  3. At day 5, the concentration equals zero, indicating complete pollutant removal, which marks the end of the monitoring period for this particular study.
  4. At day 5, the concentration becomes constant, representing equilibrium where pollutant input exactly balances natural degradation processes in the lake ecosystem.
Explanation: At t = 5, the denominator t² - 25 = 0, creating a vertical asymptote where P(t) approaches ±∞. In environmental terms, this suggests the model breaks down and concentration spikes uncontrollably. Choice B is wrong because the function approaches infinity, not negative values. Choice C is incorrect because the function is undefined, not zero. Choice D is wrong because the function doesn't approach a constant value.

Question 11

Consider the rational function g(x)=2x2−8xx2−6x+8g(x) = \frac{2x^2 - 8x}{x^2 - 6x + 8}. A student claims that since both the numerator and denominator equal zero when x=4x = 4, this creates a removable discontinuity. Which analysis of the student's reasoning is most accurate?

  1. The student is correct; factoring shows a common factor of (x - 4) that cancels, creating a hole at x = 4 that can be filled using limit evaluation. (correct answer)
  2. The student is incorrect; while x = 4 makes both parts zero, factoring reveals no common factors, so this creates a vertical asymptote instead of a removable discontinuity.
  3. The student is partially correct; x = 4 does create a discontinuity, but additional analysis shows that x = 2 also creates a removable discontinuity that the student missed.
  4. The student is incorrect; x = 4 makes the numerator zero but not the denominator, so the function equals zero at this point rather than being discontinuous.
Explanation: Factoring: g(x) = 2x(x-4)/((x-2)(x-4)). The factor (x-4) appears in both numerator and denominator, so it cancels, creating a removable discontinuity at x = 4. The limit as x approaches 4 exists. Choice B is wrong because there is a common factor. Choice C incorrectly suggests x = 2 is removable when it creates a vertical asymptote. Choice D is factually wrong about what makes the denominator zero.

Question 12

A pharmacist uses the model D(t)=250tt2+t−12D(t) = \frac{250t}{t^2 + t - 12} to represent the concentration of a drug in the bloodstream, where D(t)D(t) is concentration in mg/L and tt is hours after administration.

The model has domain restrictions that affect its medical interpretation. Which statement best explains why the domain restriction at t=3t = 3 is problematic for this pharmaceutical application?

  1. At t = 3 hours, the concentration reaches steady state, indicating optimal therapeutic levels where drug input balances metabolic elimination processes in the human body.
  2. At t = 3 hours, the concentration drops to zero, showing complete drug elimination, which means the medication is no longer effective and requires redosing.
  3. At t = 3 hours, the concentration becomes negative, representing an impossible biological scenario where the drug concentration cannot be measured meaningfully in living tissue.
  4. At t = 3 hours, the concentration becomes infinite, indicating a dangerous drug accumulation that could cause toxicity and requires immediate medical intervention to prevent overdose. (correct answer)
Explanation: When analyzing rational function models in applied contexts, you need to identify where the function is undefined and interpret what those restrictions mean in the real-world scenario. To find the domain restrictions, set the denominator equal to zero: t2+t−12=0t^2 + t - 12 = 0. Factoring gives us (t+4)(t−3)=0(t + 4)(t - 3) = 0, so t=−4t = -4 and t=3t = 3. Since time cannot be negative in this context, the critical restriction is at t=3t = 3 hours. At t=3t = 3, the denominator equals zero while the numerator equals 250(3)=750250(3) = 750. This creates the indeterminate form 7500\frac{750}{0}, which means the function approaches infinity. In pharmaceutical terms, this represents an impossible scenario where drug concentration would become infinitely large—a mathematical artifact that indicates the model breaks down at this point. Looking at the wrong answers: A incorrectly suggests steady state occurs at the discontinuity, but steady state would show constant, finite concentration. B claims concentration drops to zero, but that would require the numerator to be zero, not the denominator. C suggests negative concentration, but the function approaches positive infinity, not negative values. Answer D correctly identifies that infinite concentration represents dangerous accumulation requiring medical intervention. Study tip: When evaluating rational function models, always check where denominators equal zero and consider whether infinite values make sense in the real-world context. In biological applications, infinite quantities typically indicate model failure rather than realistic scenarios.

Question 13

The rational function k(x)=x3+2x2−3xx3+x2−6xk(x) = \frac{x^3 + 2x^2 - 3x}{x^3 + x^2 - 6x} appears to have the same factors in both numerator and denominator. After complete factorization and simplification, what is the correct description of the resulting discontinuities?

  1. Removable discontinuities at x = 0, x = 2, and x = -3, because complete factorization shows that all factors appear in both numerator and denominator and cancel out.
  2. Vertical asymptotes at x = 0, x = 2, and x = -3, because all three values make the original denominator zero and must be excluded from the domain.
  3. Removable discontinuity at x = 0 only, with vertical asymptotes at x = 2 and x = -3, because only the x factor is common to both numerator and denominator.
  4. Removable discontinuities at x = 0 and x = -3, with vertical asymptote at x = 2, because two factors cancel completely while one remains in the denominator only. (correct answer)
Explanation: When you encounter rational functions with potential common factors, your first step should be complete factorization of both numerator and denominator to identify which discontinuities are removable versus which create vertical asymptotes. Let's factor both parts of k(x)=x3+2x2−3xx3+x2−6xk(x) = \frac{x^3 + 2x^2 - 3x}{x^3 + x^2 - 6x}: Numerator: x3+2x2−3x=x(x2+2x−3)=x(x+3)(x−1)x^3 + 2x^2 - 3x = x(x^2 + 2x - 3) = x(x + 3)(x - 1) Denominator: x3+x2−6x=x(x2+x−6)=x(x+3)(x−2)x^3 + x^2 - 6x = x(x^2 + x - 6) = x(x + 3)(x - 2) So k(x)=x(x+3)(x−1)x(x+3)(x−2)k(x) = \frac{x(x + 3)(x - 1)}{x(x + 3)(x - 2)} The factors xx and (x+3)(x + 3) appear in both numerator and denominator, so they cancel out, creating removable discontinuities at x=0x = 0 and x=−3x = -3. The factor (x−2)(x - 2) remains only in the denominator, creating a vertical asymptote at x=2x = 2. Choice A incorrectly states all factors cancel—(x−2)(x - 2) doesn't appear in the numerator. Choice B treats all zeros as vertical asymptotes, ignoring that common factors create removable discontinuities. Choice C correctly identifies the removable discontinuity at x=0x = 0 but misses that (x+3)(x + 3) also cancels, making x=−3x = -3 removable, not a vertical asymptote. The correct answer is D: removable discontinuities at x=0x = 0 and x=−3x = -3, with a vertical asymptote at x=2x = 2. Study tip: Always factor completely first, then cancel common factors. Removable discontinuities occur where factors cancel; vertical asymptotes occur where only the denominator has factors.

Question 14

An economist models the relationship between tax rate and government revenue as R(x)=100x(1−x)x2−0.6x+0.09R(x) = \frac{100x(1-x)}{x^2 - 0.6x + 0.09}, where R(x)R(x) is revenue in billions and xx is the tax rate as a decimal.

The model has a domain restriction that creates economic interpretation challenges. Which statement best explains the significance of the discontinuity at x=0.3x = 0.3 in this economic context?

  1. At a 30% tax rate, the revenue equals zero, representing the optimal point where tax collection costs exactly balance the income generated from taxation policies.
  2. At a 30% tax rate, the revenue model predicts infinite income, suggesting an unrealistic economic scenario where the relationship breaks down due to taxpayer behavior changes. (correct answer)
  3. At a 30% tax rate, the revenue becomes negative, indicating that collection costs exceed income and the government loses money by implementing this particular tax level.
  4. At a 30% tax rate, the revenue reaches maximum efficiency, representing the equilibrium point where economic theory predicts optimal balance between taxation and growth.
Explanation: When analyzing rational functions in economic contexts, you need to identify discontinuities and understand their real-world implications. The denominator x2−0.6x+0.09x^2 - 0.6x + 0.09 factors as (x−0.3)2(x - 0.3)^2, creating a vertical asymptote at x=0.3x = 0.3. At this discontinuity, the function approaches infinity, meaning the revenue model predicts unlimited government income at a 30% tax rate. This is clearly unrealistic—no tax rate generates infinite revenue. In economics, such mathematical anomalies typically indicate where a model breaks down due to behavioral factors not captured in the equation. Real taxpayers would likely change behavior (evade taxes, reduce economic activity, relocate) before allowing infinite revenue extraction. Choice A incorrectly states revenue equals zero at x=0.3x = 0.3. The function is undefined there, not zero. Choice C suggests negative revenue, but the numerator 100x(1−x)100x(1-x) is positive for 0<x<10 < x < 1, so revenue approaches positive infinity, not negative values. Choice D mischaracterizes the discontinuity as an optimal equilibrium point, when it actually represents a mathematical breakdown where the model fails to reflect economic reality. The correct answer is B because it recognizes that infinite revenue predictions signal model failure due to unaccounted behavioral changes. Study tip: When analyzing rational functions in applied contexts, always check the denominator for zeros—these create discontinuities that often represent points where the mathematical model no longer reflects real-world constraints or behaviors.

Question 15

The function h(x)=x3−x2−6xx2−4x+3h(x) = \frac{x^3 - x^2 - 6x}{x^2 - 4x + 3} has multiple domain restrictions. When analyzing the discontinuities, which statement correctly identifies both the locations and types of discontinuities?

  1. Removable discontinuity at x = 0 and vertical asymptote at x = 3, because the numerator has a factor that cancels with part of the denominator.
  2. Vertical asymptotes at x = 1 and x = 3, because both values make the denominator zero and create undefined points regardless of numerator behavior.
  3. Removable discontinuity at x = 3 and vertical asymptote at x = 1, because factoring eliminates one restriction while preserving the other in the simplified form. (correct answer)
  4. Removable discontinuities at x = 1 and x = 3, because the numerator and denominator share common factors that eliminate both domain restrictions through cancellation.
Explanation: When analyzing rational functions for discontinuities, you need to factor both the numerator and denominator completely, then determine whether any zeros of the denominator are "cancelled out" by corresponding zeros in the numerator. Let's factor this function step by step. The numerator x3−x2−6x=x(x2−x−6)=x(x−3)(x+2)x^3 - x^2 - 6x = x(x^2 - x - 6) = x(x-3)(x+2). The denominator x2−4x+3=(x−1)(x−3)x^2 - 4x + 3 = (x-1)(x-3). So we have: h(x)=x(x−3)(x+2)(x−1)(x−3)h(x) = \frac{x(x-3)(x+2)}{(x-1)(x-3)} The denominator equals zero when x=1x = 1 or x=3x = 3, creating potential discontinuities at both points. However, since (x−3)(x-3) appears in both numerator and denominator, it cancels out, giving us the simplified form: h(x)=x(x+2)x−1h(x) = \frac{x(x+2)}{x-1} for x≠3x \neq 3 At x=3x = 3, we have a removable discontinuity because the common factor eliminates this restriction in the simplified function. At x=1x = 1, we have a vertical asymptote because this zero remains in the denominator after cancellation. Answer A incorrectly identifies x=0x = 0 as a discontinuity, but zero only appears in the numerator. Answer B misses that the (x−3)(x-3) factor cancels, making x=3x = 3 removable rather than a vertical asymptote. Answer D incorrectly claims both discontinuities are removable, but x=1x = 1 creates a vertical asymptote since it doesn't cancel. Remember: when factors cancel between numerator and denominator, they create removable discontinuities; when they don't cancel, they create vertical asymptotes.

Question 16

Consider m(x)=2x3−2x2−12xx3−3x2−4x+12m(x) = \frac{2x^3 - 2x^2 - 12x}{x^3 - 3x^2 - 4x + 12}. A student factors the numerator as 2x(x2−x−6)=2x(x−3)(x+2)2x(x^2 - x - 6) = 2x(x-3)(x+2) and the denominator as (x−3)(x2−4)=(x−3)(x−2)(x+2)(x-3)(x^2-4) = (x-3)(x-2)(x+2). Based on this factorization, which conclusion about discontinuities is correct?

  1. Removable discontinuity at x = 3 only, with vertical asymptotes at x = 2 and x = -2, because only the (x-3) factor appears in both numerator and denominator.
  2. Removable discontinuities at x = 3 and x = -2, with vertical asymptote at x = 2, because two factors cancel while one remains only in the denominator. (correct answer)
  3. Vertical asymptotes at x = 2, x = 3, and x = -2, because these values all make the original denominator zero and create undefined points in the function.
  4. Removable discontinuity at x = 0, with vertical asymptotes at x = 2, x = 3, and x = -2, because the factor 2x in the numerator creates additional restrictions.
Explanation: When analyzing rational functions for discontinuities, you need to identify where factors cancel versus where they create vertical asymptotes. The key is comparing the factored forms of the numerator and denominator. From the given factorization, m(x)=2x(x−3)(x+2)(x−3)(x−2)(x+2)m(x) = \frac{2x(x-3)(x+2)}{(x-3)(x-2)(x+2)}. Both the (x−3)(x-3) and (x+2)(x+2) factors appear in both numerator and denominator, so they cancel out, leaving m(x)=2xx−2m(x) = \frac{2x}{x-2} for x≠3,−2x \neq 3, -2. When factors cancel, they create removable discontinuities (holes) at x=3x = 3 and x=−2x = -2. The remaining factor (x−2)(x-2) appears only in the denominator, creating a vertical asymptote at x=2x = 2. This confirms answer B is correct. Answer A incorrectly claims only (x−3)(x-3) cancels, missing that (x+2)(x+2) also appears in both numerator and denominator. It also wrongly places a vertical asymptote at x=−2x = -2 where there's actually a removable discontinuity. Answer C assumes all zeros of the denominator create vertical asymptotes, ignoring the crucial step of canceling common factors. This is a common misconception. Answer D focuses on the factor 2x2x in the numerator, but since 2x2x doesn't appear in the denominator, it doesn't create any discontinuity at x=0x = 0. The function is actually defined and continuous there. Remember: always cancel common factors first when analyzing rational functions. Canceled factors create holes, while remaining denominator factors create vertical asymptotes.

Question 17

The rational function f(x)=x2−9x2+2x−15f(x) = \frac{x^2 - 9}{x^2 + 2x - 15} has domain restrictions that create discontinuities. After factoring completely, which statement correctly describes the nature of these discontinuities and their graphical behavior?

  1. There is a removable discontinuity at x = 3 and a vertical asymptote at x = -5, because one factor cancels while the other creates an undefined point. (correct answer)
  2. There are vertical asymptotes at both x = 3 and x = -5, because both values make the original denominator equal to zero regardless of factoring.
  3. There is a removable discontinuity at x = -3 and a vertical asymptote at x = 5, because the function simplifies to remove one restriction while maintaining the other.
  4. There are removable discontinuities at both x = 3 and x = -5, because factoring shows that both restrictions can be eliminated through algebraic simplification.
Explanation: Factoring gives: f(x) = (x-3)(x+3)/((x+5)(x-3)). The factor (x-3) cancels, creating a removable discontinuity (hole) at x = 3. The factor (x+5) remains in the denominator, creating a vertical asymptote at x = -5. Choice B ignores the cancellation. Choice C has the wrong values (signs are incorrect). Choice D incorrectly claims both are removable when x = -5 creates a vertical asymptote.

Question 18

The function f(x) = (x⁴ - 16)/(x³ + 2x² - 8x) has several domain restrictions. Which statement correctly analyzes the relationship between the zeros of the numerator and denominator?

  1. Common zeros at x = 2 and x = -2 create removable discontinuities, while x = 0 creates a single vertical asymptote
  2. No common zeros exist between numerator and denominator, so all domain restrictions create vertical asymptotes at every point
  3. Common zeros at x = 4 and x = -4 create removable discontinuities, while x = 0 and x = 2 create vertical asymptotes
  4. A common zero at x = 2 creates a removable discontinuity, while x = 0 and x = -4 create vertical asymptotes (correct answer)
Explanation: When analyzing rational functions with domain restrictions, you need to identify where the denominator equals zero, then determine whether these restrictions create removable discontinuities (holes) or vertical asymptotes based on whether the numerator also equals zero at those points. First, factor both parts of f(x)=x4−16x3+2x2−8xf(x) = \frac{x^4 - 16}{x^3 + 2x^2 - 8x}. The numerator factors as (x2−4)(x2+4)=(x−2)(x+2)(x2+4)(x^2 - 4)(x^2 + 4) = (x-2)(x+2)(x^2 + 4), giving zeros at x=2x = 2 and x=−2x = -2. The denominator factors as x(x2+2x−8)=x(x+4)(x−2)x(x^2 + 2x - 8) = x(x+4)(x-2), giving zeros at x=0x = 0, x=−4x = -4, and x=2x = 2. Since both numerator and denominator equal zero at x=2x = 2, this common zero creates a removable discontinuity (hole). At x=0x = 0 and x=−4x = -4, only the denominator equals zero, so these create vertical asymptotes. This confirms answer D. Answer A incorrectly claims x=−2x = -2 is a common zero, but x=−2x = -2 makes the denominator (−2)(2)(−4)=16≠0(-2)(2)(-4) = 16 \neq 0. Answer B wrongly states there are no common zeros, missing the shared factor (x−2)(x-2). Answer C mistakenly identifies x=4x = 4 and x=−4x = -4 as common zeros, but x=4x = 4 doesn't zero either expression, and x=−4x = -4 only zeros the denominator. Remember: removable discontinuities occur when both numerator and denominator share a common factor, while vertical asymptotes occur when only the denominator equals zero.

Question 19

Consider the function g(x) = (x³ - 8)/(x² - 4). Which statement correctly describes both the domain restrictions and the nature of the discontinuities?

  1. Domain restrictions at x = ±2; both create vertical asymptotes since the numerator is non-zero at these points
  2. Domain restrictions at x = 2 only; x = 2 creates a removable discontinuity after factoring and simplification
  3. Domain restrictions at x = ±2; x = -2 creates a removable discontinuity while x = 2 creates a vertical asymptote
  4. Domain restrictions at x = ±2; x = 2 creates a removable discontinuity while x = -2 creates a vertical asymptote (correct answer)
Explanation: When analyzing rational functions for domain restrictions and discontinuities, you need to identify where the denominator equals zero, then determine whether each restriction creates a removable discontinuity (hole) or a vertical asymptote based on whether the numerator also equals zero at that point. For g(x)=x3−8x2−4g(x) = \frac{x^3 - 8}{x^2 - 4}, start by finding where the denominator equals zero: x2−4=0x^2 - 4 = 0, so x=±2x = \pm 2. Both values are domain restrictions. Next, factor both numerator and denominator. The numerator x3−8x^3 - 8 is a difference of cubes: (x−2)(x2+2x+4)(x-2)(x^2+2x+4). The denominator factors as (x−2)(x+2)(x-2)(x+2). This gives us: g(x)=(x−2)(x2+2x+4)(x−2)(x+2)g(x) = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} At x=2x = 2: Both numerator and denominator equal zero, and the factor (x−2)(x-2) cancels out. This creates a removable discontinuity (hole). At x=−2x = -2: The denominator equals zero, but the numerator equals (−2)3−8=−16≠0(-2)^3 - 8 = -16 \neq 0. Since there's no common factor to cancel, this creates a vertical asymptote. Choice A incorrectly claims both points create vertical asymptotes. Choice B misses the restriction at x=−2x = -2 entirely. Choice C reverses which point creates which type of discontinuity. Only choice D correctly identifies that x=2x = 2 creates a removable discontinuity while x=−2x = -2 creates a vertical asymptote. Study tip: Always factor completely and check if common factors cancel—cancellation means a hole, while remaining zero denominators mean vertical asymptotes.

Question 20

A student graphs the function g(x) = (x³ - 27)/(x² - 9) and observes that the graph has a hole at one point and a vertical asymptote at another. Which analysis correctly explains these observations?

  1. A hole at x = 3 due to common factor (x - 3) and a vertical asymptote at x = -3 where only the denominator is zero (correct answer)
  2. A hole at x = -3 due to common factor (x + 3) and a vertical asymptote at x = 3 where only the denominator is zero
  3. Holes at both x = 3 and x = -3 because the numerator and denominator share factors at both points
  4. A vertical asymptote at x = 3 and another at x = -3 because both points make the denominator zero
Explanation: Factor the polynomials: numerator x³ - 27 = (x - 3)(x² + 3x + 9), denominator x² - 9 = (x - 3)(x + 3). Domain restrictions occur at x = 3 and x = -3. At x = 3, both polynomials have the common factor (x - 3), creating a removable discontinuity (hole). At x = -3, only the denominator has factor (x + 3) while the numerator (-3)³ - 27 = -27 - 27 = -54 ≠ 0, creating a vertical asymptote.