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Algebra 2

Algebra 2 Practice Test: Practice Test 6

Practice Test 6 for Algebra 2: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 has discriminant Δ=36\Delta = 36Δ=36. If the equation is solved using the quadratic formula, and the solutions are x=−4±610x = \frac{-4 \pm 6}{10}x=10−4±6​, what is the original equation?

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Question 1

A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 has discriminant Δ=36\Delta = 36Δ=36. If the equation is solved using the quadratic formula, and the solutions are x=−4±610x = \frac{-4 \pm 6}{10}x=10−4±6​, what is the original equation?

  1. 5x2−4x−1=05x^2 - 4x - 1 = 05x2−4x−1=0
  2. 10x2+8x−2=010x^2 + 8x - 2 = 010x2+8x−2=0
  3. 5x2+4x−1=05x^2 + 4x - 1 = 05x2+4x−1=0 (correct answer)
  4. 10x2−8x+2=010x^2 - 8x + 2 = 010x2−8x+2=0

Explanation: When you're given solutions from the quadratic formula and need to work backwards to find the original equation, you're essentially reverse-engineering the quadratic formula process. The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​. Comparing this to the given solutions x=−4±610x = \frac{-4 \pm 6}{10}x=10−4±6​, you can identify the components: the numerator tells us that −b=−4-b = -4−b=−4 (so b=4b = 4b=4) and Δ=36=6\sqrt{\Delta} = \sqrt{36} = 6Δ​=36​=6. The denominator shows us that 2a=102a = 102a=10, so a=5a = 5a=5. Now you need to find ccc. Since the discriminant Δ=b2−4ac=36\Delta = b^2 - 4ac = 36Δ=b2−4ac=36, substitute the known values: 42−4(5)c=364^2 - 4(5)c = 3642−4(5)c=36. This gives us 16−20c=3616 - 20c = 3616−20c=36, so −20c=20-20c = 20−20c=20, and c=−1c = -1c=−1. Therefore, the original equation is 5x2+4x−1=05x^2 + 4x - 1 = 05x2+4x−1=0, which is choice C. Let's check why the other options are wrong: Choice A has b=−4b = -4b=−4 instead of b=4b = 4b=4, which would give solutions with +4+4+4 in the numerator, not −4-4−4. Choice B has a=10a = 10a=10, making the denominator 2a=202a = 202a=20, not 10. Choice D has both b=−8b = -8b=−8 (wrong sign and magnitude) and c=2c = 2c=2 (wrong sign), leading to completely different solutions. Strategy tip: When working backwards from quadratic formula solutions, always match the pattern −b±Δ2a\frac{-b \pm \sqrt{\Delta}}{2a}2a−b±Δ​​ term by term. The denominator gives you aaa, the first part of the numerator gives you bbb, and use the discriminant formula to find ccc.

Question 2

The midpoint of the segment from z1z_1z1​ to z2z_2z2​ on the complex plane is z1+z22\dfrac{z_1+z_2}{2}2z1​+z2​​, which averages real parts and imaginary parts (just like the coordinate midpoint formula). What is the midpoint of the segment connecting z1=7−iz_1=7-iz1​=7−i and z2=1+3iz_2=1+3iz2​=1+3i?​​​

  1. 4−i4-i4−i
  2. 4+i4+i4+i (correct answer)
  3. 8+2i8+2i8+2i
  4. 3+i3+i3+i

Explanation: This question tests your understanding of finding midpoints between complex numbers on the complex plane using formulas that mirror 2D coordinate geometry. The midpoint of the segment from z₁ to z₂ is (z₁ + z₂)/2, which averages real parts and imaginary parts just like the coordinate midpoint formula! To find the midpoint of z₁ = 7 - i and z₂ = 1 + 3i: (1) Add the complex numbers: (7 - i) + (1 + 3i) = 8 + 2i. (2) Divide by 2: (8 + 2i)/2 = 8/2 + 2i/2 = 4 + i. The midpoint is at 4 + i, exactly halfway between the two points on the complex plane. Choice B correctly applies the midpoint formula to get 4 + i. Choice A (4 - i) has the wrong sign for the imaginary part—when adding -i and 3i, you get 2i, which divided by 2 gives +i, not -i. Choice C (8 + 2i) is the sum without dividing by 2—don't forget the averaging step! The midpoint formula: (1) Add both complex numbers first. (2) Then divide the entire result by 2. From (7 - i) + (1 + 3i) = 8 + 2i, divide both real and imaginary parts: 8/2 + 2i/2 = 4 + i. The complex plane formulas are identical to coordinate geometry because the complex plane IS a coordinate system!

Question 3

Based on the graph shown, which function could represent the exponential curve displayed?

  1. f(x)=2x+1f(x) = 2^x + 1f(x)=2x+1
  2. f(x)=2x+1f(x) = 2^{x+1}f(x)=2x+1
  3. f(x)=2x−1f(x) = 2^x - 1f(x)=2x−1 (correct answer)
  4. f(x)=−2x+3f(x) = -2^x + 3f(x)=−2x+3

Explanation: From the graph, the curve passes through (0, 0) and has a horizontal asymptote at y = -1. Testing the options: For choice C, f(x) = 2^x - 1: f(0) = 2^0 - 1 = 1 - 1 = 0 ‚úì, and the asymptote is y = -1 ‚úì. Choice A gives f(0) = 2 and asymptote y = 1. Choice B gives f(0) = 2 and asymptote y = 0. Choice D gives f(0) = 2 and is decreasing (reflected).

Question 4

Find the quotient q(x)q(x)q(x) and remainder r(x)r(x)r(x) when dividing x3+4x2−2x+7x^3+4x^2-2x+7x3+4x2−2x+7 by x+2x+2x+2, and write x3+4x2−2x+7x+2=q(x)+rx+2\frac{x^3+4x^2-2x+7}{x+2}=q(x)+\frac{r}{x+2}x+2x3+4x2−2x+7​=q(x)+x+2r​ with deg⁡(r)<deg⁡(x+2)\deg(r)<\deg(x+2)deg(r)<deg(x+2).

  1. x2+2x−6+19x+2x^2+2x-6+\dfrac{19}{x+2}x2+2x−6+x+219​ (correct answer)
  2. x2+2x−6−19x+2x^2+2x-6-\dfrac{19}{x+2}x2+2x−6−x+219​
  3. x2+6x−14+35x+2x^2+6x-14+\dfrac{35}{x+2}x2+6x−14+x+235​
  4. x2+2x+6+−5x+2x^2+2x+6+\dfrac{-5}{x+2}x2+2x+6+x+2−5​

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Let's divide (x³ + 4x² - 2x + 7) by (x + 2): First, x³/x = x², multiply (x + 2) by x² to get x³ + 2x², subtract to get 2x² - 2x + 7. Next, 2x²/x = 2x, multiply (x + 2) by 2x to get 2x² + 4x, subtract to get -6x + 7. Then, -6x/x = -6, multiply (x + 2) by -6 to get -6x - 12, subtract to get 19. Since 19 has degree 0 < degree 1 of (x + 2), we stop! Choice A correctly gives quotient x² + 2x - 6 and remainder 19, written as x² + 2x - 6 + 19/(x + 2). Choice B has the wrong sign on the remainder fraction, C has an incorrect quotient from arithmetic errors, and D has both wrong quotient and remainder. The degree requirement tells you when to stop: once your remainder (19) has degree 0, which is less than the divisor's degree 1, you're done—continuing would be like dividing 2 by 5 further after getting 0 remainder 2!

Question 5

Rewrite by viewing a repeated sub-expression as a single unit: 5(x−2)2+7(x−2)−3.5(x-2)^2 + 7(x-2) - 3.5(x−2)2+7(x−2)−3. Let u=(x−2)u = (x-2)u=(x−2). Which choice correctly describes the structure in terms of uuu (without expanding back in xxx)?

  1. It becomes 5u2+7u−35u^2 + 7u - 35u2+7u−3. (correct answer)
  2. It becomes (5u+7)2−3(5u+7)^2 - 3(5u+7)2−3.
  3. It becomes 5u+7u−3u25u + 7u - 3u^25u+7u−3u2.
  4. It becomes 5(x−2)2+7u−35(x-2)^2 + 7u - 35(x−2)2+7u−3, because only one (x−2)(x-2)(x−2) can be replaced.

Explanation: This question tests your ability to interpret complicated expressions by viewing one or more parts as a single entity—a powerful technique for understanding structure and relationships in complex algebraic forms. When expressions repeat sub-parts, chunking by substitution reveals hidden patterns: for example, in 5(x−2)2+7(x−2)−35(x-2)^2 + 7(x-2) - 35(x−2)2+7(x−2)−3, letting u=(x−2)u = (x-2)u=(x−2) transforms it into 5u2+7u−35u^2 + 7u - 35u2+7u−3, showing it's a quadratic in uuu. The key insight: every (x−2)(x-2)(x−2) is replaced consistently, simplifying the structure without expanding! This substitution highlights the expression's polynomial nature in terms of uuu. Viewing repeated parts as a single variable makes complex expressions easier to analyze. In the expression 5(x−2)2+7(x−2)−35(x-2)^2 + 7(x-2) - 35(x−2)2+7(x−2)−3, let's view (x−2)(x-2)(x−2) as a single entity—call it uuu. Then the expression becomes 5u2+7u−35u^2 + 7u - 35u2+7u−3, a quadratic polynomial in uuu. Now we can see: (1) The structure is quadratic with coefficients 5, 7, -3, (2) No terms are left in xxx because all (x−2)(x-2)(x−2) are replaced, (3) This reveals the expression is quadratic overall since uuu is linear in xxx. This chunking simplifies analysis without full expansion. Understanding this helps in completing the square or factoring! Choice A correctly interprets the expression by viewing the repeated sub-expression as a single entity and rewriting it fully in terms of uuu, revealing the quadratic structure. Choice C misses replacing both instances, leaving one in (x−2)(x-2)(x−2), but chunking requires consistent substitution for all identical parts. Check: if only one is replaced, it doesn't simplify properly—both must become uuu! Chunking strategy for complicated expressions: (1) Look for products—expressions where things are multiplied together often benefit from viewing each factor as a unit, (2) Identify which variables appear where—if variable xxx appears in one factor but not another, those factors are independent regarding xxx, (3) Use substitution mentally: imagine replacing a complicated sub-expression with a single letter (like let u=u = u= expression), does this simplify the structure?, (4) Ask: if I change one variable, which parts of the expression change? This reveals dependencies. For repeated sub-expressions like (x−2)(x-2)(x−2): substituting uuu reveals polynomial degree. Independence means viewing parts separately is valid! Common chunking patterns: In a(expression)powera(\text{expression})^\text{power}a(expression)power: view (expression)power(\text{expression})^\text{power}(expression)power as single factor independent of aaa. In (polynomial)(\text{polynomial})(polynomial) divided by (polynomial)(\text{polynomial})(polynomial): view numerator and denominator as separate entities. In sum of similar terms like 5(x+1)2−3(x+1)5(x + 1)^2 - 3(x + 1)5(x+1)2−3(x+1): view (x+1)(x + 1)(x+1) as single unit (substitute u=x+1u = x + 1u=x+1 gives 5u2−3u5u^2 - 3u5u2−3u, revealing quadratic structure). In nested function f(g(x))f(g(x))f(g(x)): view g(x)g(x)g(x) as the input entity to outer function fff. Chunking isn't arbitrary—chunk in ways that reveal structure, independence, or simplify analysis!

Question 6

Use logarithms to solve 15⋅22t=50015\cdot 2^{2t}=50015⋅22t=500 for ttt. Express the solution as a logarithm (exact form) and then approximate using a calculator.

  1. t=log⁡2(1003)2≈2.529t=\dfrac{\log_2\left(\dfrac{100}{3}\right)}{2}\approx 2.529t=2log2​(3100​)​≈2.529 (correct answer)
  2. t=log⁡2(50015)2≈1.265t=\dfrac{\log_2\left(\dfrac{500}{15}\right)}{2}\approx 1.265t=2log2​(15500​)​≈1.265
  3. t=log⁡(1003)2≈0.761t=\dfrac{\log\left(\dfrac{100}{3}\right)}{2}\approx 0.761t=2log(3100​)​≈0.761
  4. t=2log⁡2(1003)≈10.116t=2\log_2\left(\dfrac{100}{3}\right)\approx 10.116t=2log2​(3100​)≈10.116

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 15·2^(2t) = 500 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 15 to get 2^(2t) = 500/15 = 100/3, then take log base 2 of both sides to get log₂(2^(2t)) = log₂(100/3), and using the inverse property log₂(2^(2t)) = 2t, we get 2t = log₂(100/3), so t = log₂(100/3)/2. Starting with 15·2^(2t) = 500, divide by 15 to get 2^(2t) = 100/3, then take log₂ of both sides: 2t = log₂(100/3), and divide by 2 to get t = log₂(100/3)/2. To approximate with a calculator, use change of base: log₂(100/3) = ln(100/3)/ln(2) ≈ 3.507/0.693 ≈ 5.058, so t ≈ 5.058/2 ≈ 2.529. Choice A correctly isolates the exponential, takes log base 2, and divides by 2 to get t = log₂(100/3)/2 ≈ 2.529. Choice B incorrectly keeps the original fraction 500/15 without simplifying—while 500/15 = 100/3, the simplified form makes the calculation clearer! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(100/3)? Use change of base: log₂(100/3) = ln(100/3)/ln(2) ≈ 5.058. The formula is log_b(x) = ln(x)/ln(b) for any base b.

Question 7

The function f(x)=x2−4x+3f(x)=x^2-4x+3f(x)=x2−4x+3 fails the horizontal line test on all real numbers because it is a parabola. Which restriction makes fff invertible?

  1. Domain: [2,∞)[2,\infty)[2,∞) (correct answer)
  2. Domain: (−∞,∞)(-\infty,\infty)(−∞,∞)
  3. Domain: [0,∞)[0,\infty)[0,∞)
  4. Domain: (−∞,1](-\infty,1](−∞,1]

Explanation: This question tests your understanding that some functions (like f(x) = x² -4x +3) aren't invertible on their full domain because they're not one-to-one, but we can make them invertible by restricting the domain to a region where they pass the horizontal line test. A function must be one-to-one (each output comes from exactly one input) to have an inverse, which we verify with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one. For f(x) = x² -4x +3, which is (x-2)² -1 with vertex at x=2, it's not one-to-one on all reals because f(1)=0 and f(3)=0. But if we restrict to x ≥ 2 (right of the vertex), the function is increasing from -1 to ∞, so every horizontal line above -1 crosses once—now it's invertible! The restriction creates one-to-one behavior. For even-degree polynomials like f(x) = x² or f(x) = (x - 3)² + 1, standard restrictions are to the right or left of the vertex: restrict to x ≥ h or x ≤ h where h is the vertex's x-coordinate. This makes the function monotonic (always increasing or always decreasing), which guarantees one-to-one. Here, [2,∞) ensures monotonic increase. Choice A correctly restricts the domain to [2,∞) to make it one-to-one and invertible. Choice C fails because [0,∞) spans both sides of the vertex (f(0)=3, f(4)=3), violating one-to-one. Domain restriction decision tree: (1) Graph the function (or visualize it), (2) Find where it fails horizontal line test (usually at vertex for parabolas), (3) Choose a domain making the function monotonic—either the increasing part or the decreasing part, (4) Standard choices: for x², use x ≥ 0; for (x - h)² + k, use x ≥ h or x ≤ h; for x⁴, use x ≥ 0. The restricted domain should be an interval [a, ∞) or (-∞, a] where the function is one-to-one. After restricting, you can find the inverse by solving y = (x-2)² -1 for x in terms of y, taking the positive branch since x ≥ 2. Great job spotting the vertex—you've got this!

Question 8

If z=a+biz = a + biz=a+bi where aaa and bbb are real, and z⋅z‾=25z \cdot \overline{z} = 25z⋅z=25, and z+z‾=6z + \overline{z} = 6z+z=6, what are the possible values of zzz?

  1. 3+4i3 + 4i3+4i only
  2. 3−4i3 - 4i3−4i only
  3. 3+4i3 + 4i3+4i or 3−4i3 - 4i3−4i (correct answer)
  4. 4+3i4 + 3i4+3i or 4−3i4 - 3i4−3i

Explanation: From z⋅z‾=25z \cdot \overline{z} = 25z⋅z=25, we get ∣z∣2=a2+b2=25|z|^2 = a^2 + b^2 = 25∣z∣2=a2+b2=25. From z+z‾=6z + \overline{z} = 6z+z=6, we get (a+bi)+(a−bi)=2a=6(a + bi) + (a - bi) = 2a = 6(a+bi)+(a−bi)=2a=6, so a=3a = 3a=3. Substituting into the first equation: 32+b2=253^2 + b^2 = 2532+b2=25, so 9+b2=259 + b^2 = 259+b2=25, giving b2=16b^2 = 16b2=16 and b=±4b = \pm 4b=±4. Therefore both z=3+4iz = 3 + 4iz=3+4i and z=3−4iz = 3 - 4iz=3−4i satisfy the conditions. Choice D has the values of aaa and bbb swapped.

Question 9

A movie theater sells popcorn in two sizes. Small costs 4andlargecosts4 and large costs 4andlargecosts7. Let xxx be the number of small popcorns, yyy be the number of large popcorns, and TTT be the total cost in dollars. Write an equation relating TTT, xxx, and yyy. Then, to graph the relationship between yyy and xxx when T=70T=70T=70, choose correct axes labels and a reasonable scale for 0≤x≤200\le x\le 200≤x≤20.

  1. Equation: T=4x+7yT=4x+7yT=4x+7y; With T=70T=70T=70, graph yyy versus xxx using x-axis: Small popcorns xxx (items), y-axis: Large popcorns yyy (items); scale: x 0 to 20 by 2, y 0 to 10 by 1. (correct answer)
  2. Equation: T=11xyT=11xyT=11xy; With T=70T=70T=70, graph with x-axis: Cost TTT (dollars), y-axis: Small popcorns xxx (items); scale: x 0 to 70 by 10, y 0 to 20 by 2.
  3. Equation: T=4x+7yT=4x+7yT=4x+7y; With T=70T=70T=70, graph xxx versus yyy using x-axis: Cost TTT (dollars), y-axis: Large popcorns yyy (items); scale: x 0 to 20 by 2, y 0 to 10 by 1.
  4. Equation: T=7x+4yT=7x+4yT=7x+4y; With T=70T=70T=70, graph yyy versus xxx using x-axis: Small popcorns xxx (items), y-axis: Large popcorns yyy (items); scale: x 0 to 20 by 2, y 0 to 10 by 1.

Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is 40perhourplus40 per hour plus 40perhourplus15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: C = 40h + 15n. The equation structure mirrors the context structure! Total cost T is 4 times small x plus 7 times large y, so T = 4x + 7y; for fixed T=70, graphing y versus x means x-axis Small popcorns x (items), y-axis Large popcorns y (items), with scale x 0-20 by 2, y 0-10 by 1 fitting max y= (70-0)/7=10 and x=(70-0)/4=17.5. Choice A correctly creates the equation T = 4x + 7y and sets up the graph with proper labels and scale. Choice D swaps coefficients to 7x + 4y, mismatching prices; ensure each rate matches the described item. Equation creation framework: (1) Define your variables clearly—'Let x = [exactly what it represents] in [units]'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 10

Use the table to answer the question. A school cafeteria plans to serve two lunch options with specific nutritional requirements. Option 1 costs 3.50perservingandOption2costs3.50 per serving and Option 2 costs 3.50perservingandOption2costs4.25 per serving. The cafeteria must serve exactly 400400400 lunches daily while staying within a $1500 daily food budget. Additionally, they need to meet minimum daily nutritional targets. If xxx represents servings of Option 1 and yyy represents servings of Option 2, which system best models these constraints?

  1. x+y=400x + y = 400x+y=400, 3.50x+4.25y≤15003.50x + 4.25y \leq 15003.50x+4.25y≤1500, 250x+180y≥90000250x + 180y \geq 90000250x+180y≥90000, 12x+18y≥560012x + 18y \geq 560012x+18y≥5600 (correct answer)
  2. x+y=400x + y = 400x+y=400, 3.50x+4.25y<15003.50x + 4.25y < 15003.50x+4.25y<1500, 250x+180y≥90000250x + 180y \geq 90000250x+180y≥90000, 12x+18y≥560012x + 18y \geq 560012x+18y≥5600
  3. x+y≤400x + y \leq 400x+y≤400, 3.50x+4.25y≤15003.50x + 4.25y \leq 15003.50x+4.25y≤1500, 250x+180y≥90000250x + 180y \geq 90000250x+180y≥90000, 12x+18y≥560012x + 18y \geq 560012x+18y≥5600
  4. x+y=400x + y = 400x+y=400, 3.50x+4.25y≤15003.50x + 4.25y \leq 15003.50x+4.25y≤1500, 180x+250y≥90000180x + 250y \geq 90000180x+250y≥90000, 18x+12y≥560018x + 12y \geq 560018x+12y≥5600

Explanation: The correct answer is A. The constraints are: exactly 400 lunches (x+y=400x + y = 400x+y=400), budget limit (3.50x+4.25y≤15003.50x + 4.25y \leq 15003.50x+4.25y≤1500), minimum calories (250x+180y≥90000250x + 180y \geq 90000250x+180y≥90000), and minimum protein (12x+18y≥560012x + 18y \geq 560012x+18y≥5600). Choice B incorrectly uses strict inequality for budget (should allow equality at $1500). Choice C allows serving fewer than 400 lunches when exactly 400 is required. Choice D swaps the nutritional values between options (Option 1 has 250 calories and 12g protein, not Option 2's values).

Question 11

Graph the logarithmic function f(x)=ln⁡(x)+2f(x)=\ln(x)+2f(x)=ln(x)+2 showing the vertical asymptote and the x-intercept.

  1. Vertical asymptote x=−2x=-2x=−2; x-intercept (2,0)(2,0)(2,0).
  2. Vertical asymptote x=0x=0x=0; x-intercept (e−2,0)(e^{-2},0)(e−2,0). (correct answer)
  3. Vertical asymptote x=2x=2x=2; x-intercept (0,2)(0,2)(0,2).
  4. Vertical asymptote y=2y=2y=2; x-intercept (1,0)(1,0)(1,0).

Explanation: This question tests your ability to graph logarithmic functions by identifying their characteristic features like asymptotes and intercepts. Logarithmic functions f(x) = log_b(x) are the inverses of exponentials, so their graphs are reflections across y = x: they have an x-intercept at (1, 0) because log_b(1) = 0, no y-intercept because log_b(0) is undefined, and a vertical asymptote at x = 0 (the y-axis) that the graph approaches as x → 0⁺. The domain is restricted to x > 0 (can't take log of negative or zero), and end behavior is: as x → 0⁺, f(x) → -∞ (graph goes down along the asymptote), and as x → ∞, f(x) → ∞ (graph rises slowly, flattening as it goes). For f(x) = ln(x) + 2, it's a natural log (base e) shifted up by 2, with vertical asymptote still at x=0 (no horizontal shift); x-intercept when ln(x) + 2 = 0 so ln(x) = -2, x = e^{-2}. Choice B correctly identifies the vertical asymptote as x=0 and x-intercept as (e^{-2}, 0). A distractor like Choice C might confuse vertical with horizontal asymptote, but logs have vertical asymptotes, not horizontal ones. Exponential vs logarithmic graphing comparison: exponentials have y-intercept and horizontal asymptote (HA), while logarithms have x-intercept and vertical asymptote (VA). They're mirror images across y = x! Both never cross their asymptote. For exponentials, check the base: b > 1 means rising (growth), 0 < b < 1 means falling (decay). For logarithms, the graph always rises from left to right (slowly), hugging the VA on the left and flattening as it goes right. These characteristic shapes are instantly recognizable!

Question 12

Derive the equation of a circle with center (0,0)(0,0)(0,0) and radius 333 using the idea that the distance from (x,y)(x,y)(x,y) to the center is 333.

  1. x2+y2=6x^2 + y^2 = 6x2+y2=6
  2. (x−3)2+(y−3)2=9(x-3)^2 + (y-3)^2 = 9(x−3)2+(y−3)2=9
  3. x2+y2=9x^2 + y^2 = 9x2+y2=9 (correct answer)
  4. x2+y2=3x^2 + y^2 = 3x2+y2=3

Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. A circle is defined as all points at a fixed distance (radius r) from a center point (h, k): using the distance formula, the distance from any point (x, y) on the circle to the center is (x−h)2+(y−k)2=r\sqrt{(x - h)^2 + (y - k)^2} = r(x−h)2+(y−k)2​=r. Squaring both sides eliminates the radical and gives the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2. This equation comes directly from the Pythagorean Theorem applied to the right triangle formed by the horizontal distance (x - h), vertical distance (y - k), and radius r as hypotenuse! For center (0, 0) and radius 3, it's simply x2+y2=9x^2 + y^2 = 9x2+y2=9, since (x−0)2+(y−0)2=32(x - 0)^2 + (y - 0)^2 = 3^2(x−0)2+(y−0)2=32. Choice C correctly derives the equation with r² = 9 for the origin-centered circle. A distractor like choice B might use 6 instead of 9, perhaps confusing radius with diameter, but remember to square the radius. Deriving from center and radius: (1) Write the distance from general point (x, y) to center (h, k) using distance formula: (x−h)2+(y−k)2\sqrt{(x - h)^2 + (y - k)^2}(x−h)2+(y−k)2​, (2) Set equal to radius r, (3) Square both sides to get (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2. That's it!

Question 13

Order the following functions by eventual growth rate (fastest to slowest):

  • a(x)=0.001⋅3xa(x)=0.001\cdot 3^xa(x)=0.001⋅3x (exponential)
  • b(x)=x6b(x)=x^6b(x)=x6 (polynomial)
  • c(x)=500x2c(x)=500x^2c(x)=500x2 (quadratic)
  • d(x)=12xd(x)=12xd(x)=12x (linear)​
  1. b>a>c>db>a>c>db>a>c>d
  2. a>b>c>da>b>c>da>b>c>d (correct answer)
  3. a>c>b>da>c>b>da>c>b>d
  4. c>b>a>dc>b>a>dc>b>a>d

Explanation: This question tests your understanding of a fundamental mathematical principle: exponential functions eventually grow faster than any polynomial function (even very high-degree polynomials) when we look at sufficiently large x-values. The growth hierarchy is: exponential > any polynomial > linear (for large x). Even a slow exponential like 0.001·3^x will eventually exceed a fast polynomial like x^6 if you go far enough. This happens because exponential growth is multiplicative (multiply by same factor repeatedly, which compounds), while polynomial growth is essentially additive-based (even with acceleration). Multiplicative compounding always beats any additive pattern eventually—it's why compound interest (exponential) is so powerful long-term compared to simple interest (linear)! Despite the tiny coefficient 0.001, the exponential a(x)=0.001·3^x will eventually dominate because 3^x grows so rapidly. At x=20: a(20)=0.001·3^20≈3,486,784 while b(20)=20^6=64,000,000 (polynomial still larger), but by x=30: a(30)=0.001·3^30≈205 billion while b(30)=30^6≈729 million (exponential has dominated!). Choice B correctly orders them as a>b>c>d, recognizing that the exponential eventually dominates, followed by the 6th-degree polynomial, then the quadratic (despite its large coefficient), and finally the linear function. Choice A incorrectly puts the polynomial first, not recognizing exponential dominance. Observing the hierarchy: the order is determined by the type of function (exponential > polynomial) and for polynomials, by degree (x^6 > x^2 > x). Coefficients like 0.001 or 500 only affect when crossovers occur, not the eventual ordering. Why exponential beats polynomial: 3^x triples each step (×3), while x^6 grows by adding larger amounts. Tripling repeatedly (even starting from 0.001) eventually outpaces any polynomial growth—it's the power of compound growth!

Question 14

Show the symmetry type by identifying whether f(x)=x3−xf(x)=x^3-xf(x)=x3−x is even, odd, or neither.

  1. Even
  2. Odd (correct answer)
  3. Neither
  4. Both even and odd

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. To test f(x) = x³ - x, we find f(-x) = (-x)³ - (-x) = -x³ + x = -(x³ - x) = -f(x), confirming it's odd. Choice B correctly identifies this function as odd (symmetric about the origin). The function cannot be even since f(-x) = -f(x) ≠ f(x), and it clearly satisfies the odd function definition. Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation. Most functions are neither even nor odd!

Question 15

Let f(x)=xf(x)=\sqrt{x}f(x)=x​ be the parent function. Describe the effect of g(x)=f(2x)g(x)=f(2x)g(x)=f(2x) on the graph of fff.

  1. Horizontal stretch by a factor of 2
  2. Vertical stretch by a factor of 2
  3. Horizontal compression by a factor of 2 (correct answer)
  4. Shift left 2 units

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. With g(x) = f(2x) for the square root function, multiplying x by 2 (inside) squeezes the graph horizontally toward the y-axis, making it steeper and narrower—excellent work recognizing this! Choice B correctly identifies the transformation as a horizontal compression by a factor of 2. Choice A might confuse compression with stretch, but since |2| > 1, it's compression, not stretch—you've got this! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 16

Function a(x)=3sin⁡(2x)+1a(x) = 3\sin(2x) + 1a(x)=3sin(2x)+1 and function bbb is shown in the table. Compare the periods of these two functions.

  1. Function aaa has period π\piπ and function bbb appears to have period 888
  2. Function aaa has period 2π2\pi2π and function bbb appears to have period 444
  3. Function aaa has period π\piπ and function bbb appears to have period 444 (correct answer)
  4. Both functions have the same period of π\piπ

Explanation: For a(x) = 3sin(2x) + 1, the period is 2π/|coefficient of x| = 2π/2 = π. For function b, examining the table values: b(0) = 2, b(1) = 5, b(2) = 2, b(3) = -1, b(4) = 2, b(5) = 5. We see that b(0) = b(4) = 2 and b(1) = b(5) = 5, suggesting the pattern repeats every 4 units, so the period appears to be 4. Choice A incorrectly identifies b's period as 8. Choice B incorrectly states a's period as 2π. Choice D incorrectly claims both periods are equal.

Question 17

Convert 4(cos⁡270∘+isin⁡270∘)4\left(\cos 270^\circ+i\sin 270^\circ\right)4(cos270∘+isin270∘) to rectangular form a+bia+bia+bi using a=rcos⁡θa=r\cos\thetaa=rcosθ and b=rsin⁡θb=r\sin\thetab=rsinθ.

  1. 4+0i4+0i4+0i
  2. 0+4i0+4i0+4i
  3. 0−4i0-4i0−4i (correct answer)
  4. −4+0i-4+0i−4+0i

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert FROM polar TO rectangular: use a = r cos θ and b = r sin θ. For 4(cos 270° + i sin 270°): (1) Identify r = 4 and θ = 270°. (2) Calculate real part: a = 4 cos 270° = 4(0) = 0. (3) Calculate imaginary part: b = 4 sin 270° = 4(-1) = -4. (4) Write rectangular form: 0 + (-4)i = 0 - 4i. Choice C correctly applies the conversion formulas with the special angle values cos 270° = 0 and sin 270° = -1. Choice B would be correct for 4(cos 90° + i sin 90°) = 0 + 4i, confusing 270° (pointing down on negative imaginary axis) with 90° (pointing up on positive imaginary axis). Remember: 270° points straight down! Polar to rectangular recipe: (1) Visualize angles on unit circle: 0° = right, 90° = up, 180° = left, 270° = down. (2) At 270°, we're on negative y-axis: cos 270° = 0 (no horizontal component), sin 270° = -1 (full negative vertical). (3) Multiply by r: a = 4(0) = 0, b = 4(-1) = -4. The four cardinal directions give pure real or pure imaginary numbers: 0° → r + 0i, 90° → 0 + ri, 180° → -r + 0i, 270° → 0 - ri.

Question 18

The quadratic f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c has roots r1=2+3ir_1 = 2 + 3ir1​=2+3i and r2=2−3ir_2 = 2 - 3ir2​=2−3i. If f(0)=39f(0) = 39f(0)=39, what is the value of the leading coefficient aaa?

  1. a=3a = 3a=3 (correct answer)
  2. a=13a = 13a=13
  3. a=−3a = -3a=−3
  4. a=−13a = -13a=−13

Explanation: Since the roots are 2+3i and 2-3i, we can write f(x) = a(x-(2+3i))(x-(2-3i)) = a((x-2)-3i)((x-2)+3i) = a((x-2)² + 9) = a(x² - 4x + 4 + 9) = a(x² - 4x + 13). Since f(0) = 39, we have a(13) = 39, so a = 3. Choice B incorrectly uses a = 13 from the constant term. Choice C uses a = -3 from sign error. Choice D combines both errors.

Question 19

The expression 5(x−2)2−3(x+1)(x−2)5(x - 2)^2 - 3(x + 1)(x - 2)5(x−2)2−3(x+1)(x−2) contains the factor (x−2)(x - 2)(x−2). What is the other factor when this expression is completely factored?

  1. The other factor is (5x−10−3x−3)(5x - 10 - 3x - 3)(5x−10−3x−3)
  2. The other factor is (5x−13)(5x - 13)(5x−13)
  3. The other factor is (2x−13)(2x - 13)(2x−13) (correct answer)
  4. The other factor is (5x−7)(5x - 7)(5x−7)

Explanation: Factor out (x−2)(x - 2)(x−2): 5(x−2)2−3(x+1)(x−2)=(x−2)[5(x−2)−3(x+1)]=(x−2)[5x−10−3x−3]=(x−2)(2x−13)5(x - 2)^2 - 3(x + 1)(x - 2) = (x - 2)[5(x - 2) - 3(x + 1)] = (x - 2)[5x - 10 - 3x - 3] = (x - 2)(2x - 13)5(x−2)2−3(x+1)(x−2)=(x−2)[5(x−2)−3(x+1)]=(x−2)[5x−10−3x−3]=(x−2)(2x−13). The other factor is (2x−13)(2x - 13)(2x−13). Choice A shows the unsimplified form before combining like terms. Choice B incorrectly keeps the coefficient 555 from the first term. Choice D results from an arithmetic error when combining like terms.

Question 20

A community pool has two pumps that can fill it. Pump A can fill the pool in 6 hours, and Pump B can fill the pool in 8 hours.

Write and solve a rational equation to find how long it takes to fill the pool if both pumps run at the same time.

Let ttt = the time (in hours) to fill the pool together.

  1. Set up 6+8=t6+8=t6+8=t. Then t=14t=14t=14 hours, so it takes 14 hours together.
  2. Set up 16+18=1t\frac{1}{6}+\frac{1}{8}=\frac{1}{t}61​+81​=t1​. Then t=143t=\frac{14}{3}t=314​ hours, so it takes about 4.67 hours together.
  3. Set up 16+18=1t\frac{1}{6}+\frac{1}{8}=\frac{1}{t}61​+81​=t1​. Then t=247t=\frac{24}{7}t=724​ hours, so it takes about 3.43 hours together. (correct answer)
  4. Set up 6t+8t=1\frac{6}{t}+\frac{8}{t}=1t6​+t8​=1. Then t=14t=14t=14 hours, so it takes 14 hours together.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. For this pool-filling scenario, set up the equation as 1/6 + 1/8 = 1/t; find a common denominator of 24 to get 4/24 + 3/24 = 7/24 = 1/t, so t = 24/7 ≈ 3.43 hours, meaning both pumps together fill the pool in about 3 hours 26 minutes. Choice B correctly sets up the equation by adding the rates and solves to find t = 24/7 hours, accurately determining the combined time. A common mistake, as in choice C, is adding the individual times instead of the rates, which overestimates the combined time since they work together faster. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 21

Find all points where the line intersects the circle (solve the system).

y = x \\ x^2 + y^2 = 8 \end{cases}$$
  1. (2,2)(\sqrt{2},\sqrt{2})(2​,2​) and (−2,−2)(-\sqrt{2},-\sqrt{2})(−2​,−2​)
  2. (8,8)(\sqrt{8},\sqrt{8})(8​,8​) and (−8,−8)(-\sqrt{8},-\sqrt{8})(−8​,−8​)
  3. (2,2)(2,2)(2,2) and (−2,−2)(-2,-2)(−2,−2) (correct answer)
  4. (2,−2)(\sqrt{2},-\sqrt{2})(2​,−2​) and (−2,2)(-\sqrt{2},\sqrt{2})(−2​,2​)

Explanation: This question tests your ability to solve systems of a line and circle to find all intersection points. A linear-quadratic system with a circle can have 0, 1, or 2 intersections based on the line's position. For y = x and x2+y2=8x^2 + y^2 = 8x2+y2=8, substitute: x2+x2=8x^2 + x^2 = 8x2+x2=8, 2x2=82x^2 = 82x2=8, x2=4x^2 = 4x2=4, x=±2x = \pm 2x=±2; y=±2y = \pm 2y=±2, giving (2,2)(2,2)(2,2) and (−2,−2)(-2,-2)(−2,−2), both valid. Choice A correctly identifies the points from proper substitution. Choice B might result from 8\sqrt{8}8​ instead of 4\sqrt{4}4​, but solve the quadratic accurately after combining terms! Strategy: substitute linear into circle, simplify, solve for x, find y, verify; discriminant previews the count. Wonderful effort—this will sharpen your geometry-algebra skills!

Question 22

Two savings plans are described below.

Plan 1: Start with 2000andadd2000 and add 2000andadd50 each month. Plan 2: Start with $2000 and increase the balance by 2% each month.

Which plan shows constant percent change (exponential), and which shows constant additive change (linear)?

  1. Plan 1 is exponential; Plan 2 is linear
  2. Both plans are exponential because both increase each month
  3. Both plans are linear because they both change monthly
  4. Plan 2 is exponential; Plan 1 is linear (correct answer)

Explanation: This question tests your ability to recognize exponential growth or decay—situations where a quantity changes by a constant percent (not constant amount) per time interval. Constant percent growth means multiplying by the same factor greater than 1 each interval: if something grows 5% per year, each year's value is 105% of the previous (multiply by 1.05). Constant percent decay means multiplying by a factor between 0 and 1: if something decreases 15% per year, each year retains 85% of previous (multiply by 0.85). The key test: calculate ratios of consecutive values. If ratios are constant, it's exponential with that ratio as the growth/decay factor! Plan 1 adds 50eachmonth—thisisconstantadditivechange,makingitlinear.Plan2increasesthebalanceby250 each month—this is constant additive change, making it linear. Plan 2 increases the balance by 2% each month, meaning each month's balance is 102% of the previous (multiply by 1.02)—this is constant percent change, making it exponential. Choice D correctly identifies Plan 2 as exponential (constant percent change) and Plan 1 as linear (constant additive change). Choice A reverses these classifications, missing that "add 50eachmonth—thisisconstantadditivechange,makingitlinear.Plan2increasesthebalanceby250" signals linear while "increase by 2%" signals exponential. The ratio test for exponential: (1) from table, divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, (2) if all equal → exponential with that ratio as base b, (3) if b > 1 → growth; if 0 < b < 1 → decay, (4) calculate percent rate: r = b - 1, convert to percent. Example: ratios all 1.06 → base 1.06 → growth → rate = 1.06 - 1 = 0.06 = 6% per interval. This systematic check identifies exponential patterns reliably! Don't confuse exponential with linear: linear adds the same amount each time (constant differences like +50, +50, +50), exponential multiplies by same factor (constant ratios like ×1.1, ×1.1, ×1.1). Both are patterns of regular change, but different mechanisms!

Question 23

Determine whether the system is consistent. If it has a solution, give it; otherwise state no solution:

3x - y = 2\\ 6x - 2y = 5 \end{cases}$$
  1. Infinitely many solutions
  2. No solution (correct answer)
  3. (1,1)(1, 1)(1,1)
  4. (2,4)(2, 4)(2,4)

Explanation: This question tests your ability to determine whether a system is consistent (has at least one solution) or inconsistent (has no solution), requiring analysis of the relationship between equations. A system is inconsistent when the equations represent parallel lines that never intersect, which happens when they have the same slope but different y-intercepts. Looking at 3x - y = 2 and 6x - 2y = 5, let's check if the second equation is a multiple of the first: if we multiply the first equation by 2, we get 2(3x - y) = 2(2), which gives 6x - 2y = 4. But our second equation has 6x - 2y = 5, not 4! This means the left sides are proportional but the right sides are not (4 ≠ 5), indicating parallel lines. We can verify by converting to slope-intercept form: from 3x - y = 2, we get y = 3x - 2; from 6x - 2y = 5, we get 2y = 6x - 5, so y = 3x - 5/2. Same slope (3) but different y-intercepts (-2 vs -5/2), confirming parallel lines. Choice A correctly identifies that the system has no solution because the lines are parallel—they have the same slope but different y-intercepts, so they never intersect. Choice B incorrectly assumes that having proportional left sides means infinitely many solutions, but the right sides must also be proportional for that to be true. To check consistency: (1) See if one equation is a constant multiple of the other (consistent with infinitely many solutions), (2) Check if left sides are proportional but right sides are not (inconsistent, no solution), (3) Otherwise, the system has exactly one solution. Remember: parallel lines (same slope, different y-intercepts) mean no solution, while the same line written two ways means infinitely many solutions!

Question 24

A medicine dose remaining in the bloodstream is measured every 6 hours:

Time (hours): 0, 6, 12, 18 Amount (mg): 80, 60, 45, 33.75

Is this constant percent decay? If so, what is the percent decrease per 6 hours?

  1. Yes; 25% decay per 6 hours (decay factor b=0.75b=0.75b=0.75) (correct answer)
  2. Yes; 20% decay per 6 hours (decay factor b=0.80b=0.80b=0.80)
  3. No; it is linear because it decreases by 15 mg each interval
  4. Yes; 75% decay per 6 hours (decay factor b=0.25b=0.25b=0.25)

Explanation: This question tests your ability to recognize exponential growth or decay—situations where a quantity changes by a constant percent (not constant amount) per time interval. Constant percent growth means multiplying by the same factor greater than 1 each interval: if something grows 5% per year, each year's value is 105% of the previous (multiply by 1.05). Constant percent decay means multiplying by a factor between 0 and 1: if something decreases 15% per year, each year retains 85% of previous (multiply by 0.85). The key test: calculate ratios of consecutive values. If ratios are constant, it's exponential with that ratio as the growth/decay factor! Let's check the ratios: 60/80 = 0.75, 45/60 = 0.75, 33.75/45 = 0.75. All ratios equal 0.75, confirming exponential decay with base b = 0.75. Since 0.75 = 1 - 0.25, this represents 25% decay per 6 hours. Choice A correctly identifies 25% decay per 6 hours (decay factor b=0.75) through the constant ratio of 0.75. Choice C incorrectly suggests linear decay by looking at differences (80-60=20, not 15), and fails to check that differences aren't even constant. The ratio test for exponential: (1) from table, divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, (2) if all equal → exponential with that ratio as base b, (3) if b > 1 → growth; if 0 < b < 1 → decay, (4) calculate percent rate: r = b - 1, convert to percent. Example: ratios all 1.06 → base 1.06 → growth → rate = 1.06 - 1 = 0.06 = 6% per interval. This systematic check identifies exponential patterns reliably! Don't confuse exponential with linear: linear adds the same amount each time (constant differences like +50, +50, +50), exponential multiplies by same factor (constant ratios like ×1.1, ×1.1, ×1.1). Both are patterns of regular change, but different mechanisms!

Question 25

A medication amount in the bloodstream is modeled by M(t)=80(0.92)tM(t)=80(0.92)^tM(t)=80(0.92)t, where ttt is in hours.

What percent does the amount change each hour, and is it growth or decay?​

  1. 8% decay per hour (correct answer)
  2. 92% decay per hour
  3. 8% growth per hour
  4. 0.92% decay per hour

Explanation: This question tests your ability to recognize exponential growth or decay—situations where a quantity changes by a constant percent (not constant amount) per time interval. Constant percent growth means multiplying by the same factor greater than 1 each interval: if something grows 5% per year, each year's value is 105% of the previous (multiply by 1.05). Constant percent decay means multiplying by a factor between 0 and 1: if something decreases 15% per year, each year retains 85% of previous (multiply by 0.85). The key test: calculate ratios of consecutive values. If ratios are constant, it's exponential with that ratio as the growth/decay factor! From the function M(t) = 80(0.92)^t, the base is 0.92, indicating a constant multiplicative factor less than 1 each hour. Choice A correctly identifies 8% decay per hour through the base analysis (0.92 = 1 - 0.08). A distractor like choice B misinterprets the base as the percent decay directly, but it's the retention factor—remember, percent decay is 1 - base! The ratio test for exponential: (1) from table, divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, (2) if all equal → exponential with that ratio as base b, (3) if b > 1 → growth; if 0 < b < 1 → decay, (4) calculate percent rate: r = b - 1, convert to percent. Example: ratios all 1.06 → base 1.06 → growth → rate = 1.06 - 1 = 0.06 = 6% per interval. This systematic check identifies exponential patterns reliably! Don't confuse exponential with linear: linear adds the same amount each time (constant differences like +50, +50, +50), exponential multiplies by same factor (constant ratios like ×1.1, ×1.1, ×1.1). Both are patterns of regular change, but different mechanisms! Check both: if differences constant → linear. If ratios constant → exponential. Usually only one pattern holds. The language helps too: 'grows by $50 per year' = linear (additive), 'grows by 5% per year' = exponential (multiplicative)!