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Algebra 2

Algebra 2 Practice Test: Practice Test 101

Practice Test 101 for Algebra 2: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A rectangular patio has an area of 96 square meters. Its length is 4 meters more than its width.

Write and solve an equation to find the width and length.

Let www = the width in meters.

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Question 1

A rectangular patio has an area of 96 square meters. Its length is 4 meters more than its width.

Write and solve an equation to find the width and length.

Let www = the width in meters.

  1. Set up w(w+4)=96w(w+4)=96w(w+4)=96. Then w=8w=8w=8 (reject w=−12w=-12w=−12). Width 8 m and length 12 m. (correct answer)
  2. Set up w+w+4=96w+w+4=96w+w+4=96. Then w=46w=46w=46. Width 46 m and length 50 m.
  3. Set up w2+4=96w^2+4=96w2+4=96. Then w=92w=\sqrt{92}w=92​. Width 92\sqrt{92}92​ m and length 92+4\sqrt{92}+492​+4 m.
  4. Set up w(w−4)=96w(w-4)=96w(w−4)=96. Then w=12w=12w=12 (reject w=−8w=-8w=−8). Width 12 m and length 8 m.

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. Area problems with relationships between dimensions often lead to quadratic equations: if length is width plus a constant, area = width × (width + constant), forming a quadratic like w(w + 4) = 96 that you solve using factoring or quadratic formula, rejecting negative roots. Here, the equation is w(w + 4) = 96, expanding to w² + 4w - 96 = 0; solving gives w = [-4 ± √(16 + 384)]/2 = [-4 ± √400]/2 = [-4 ± 20]/2, so w = 8 or w = -12 (rejected), with length 12 m. Choice A correctly sets up the quadratic from area and solves to find width 8 m and length 12 m. A distractor like choice B confuses area with perimeter, leading to incorrect large dimensions that don't match the given area. 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 2

Solve for xxx: log⁡4(x)=0\log_4(x)=0log4​(x)=0.

  1. x=0x=0x=0
  2. x=1x=1x=1 (correct answer)
  3. x=4x=4x=4
  4. x=−1x=-1x=−1

Explanation: This question tests your understanding that logarithms and exponentials are inverse operations—logarithms undo exponentiation and vice versa, just like square roots undo squaring. The fundamental connection: log_b(x) = y means exactly the same thing as b^y = x. The logarithm log_b(x) asks 'what power of b gives x?', and the answer is that exponent y. Given log₄(x) = 0, we convert to exponential form: 4⁰ = x. Since any non-zero number raised to the power 0 equals 1, we have 4⁰ = 1, so x = 1. Choice B correctly identifies that x = 1. Choice A might tempt you since 0 appears in the equation, but remember that 0 is the exponent, not the result—and 4⁰ = 1, not 0. A key fact to remember: log_b(1) = 0 for any valid base b, because b⁰ = 1. This makes sense with the inverse relationship—to get 1, you raise any base to the power 0!

Question 3

A phone battery is at 90% at the start of the day and then retains 80% of its charge at the end of each hour (due to constant use). Let ana_nan​ be the battery percentage after n−1n-1n−1 hours (so a1=90a_1=90a1​=90), with domain n∈{1,2,3,… }n\in\{1,2,3,\dots\}n∈{1,2,3,…}. Write both a recursive definition and an explicit formula.

  1. Recursive: a1=90, an+1=an−20a_1=90,\ a_{n+1}=a_n-20a1​=90, an+1​=an​−20; Explicit: an=90−20(n−1)a_n=90-20(n-1)an​=90−20(n−1)
  2. Recursive: a1=90, an+1=0.8ana_1=90,\ a_{n+1}=0.8a_na1​=90, an+1​=0.8an​; Explicit: an=90⋅(0.8)n−1a_n=90\cdot(0.8)^{n-1}an​=90⋅(0.8)n−1 (correct answer)
  3. Recursive: a1=90, an+1=an⋅90a_1=90,\ a_{n+1}=a_n\cdot 90a1​=90, an+1​=an​⋅90; Explicit: an=0.8⋅90n−1a_n=0.8\cdot 90^{n-1}an​=0.8⋅90n−1
  4. Recursive: a1=72, an+1=0.8ana_1=72,\ a_{n+1}=0.8a_na1​=72, an+1​=0.8an​; Explicit: an=72⋅(0.8)n−1a_n=72\cdot(0.8)^{n-1}an​=72⋅(0.8)n−1

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! For the battery model starting at 90% and retaining 80% (r=0.8) each hour, it's geometric with a₁=90, so recursive a₁=90, aₙ₊₁=0.8aₙ and explicit aₙ=90·(0.8)^{n-1}. Choice B correctly writes both formulas, modeling the exponential decay. Choices like A use arithmetic subtraction, but multiplication by 0.8 makes it geometric—fantastic application to real life! Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 4

A science sensor records the temperature change according to T(t)=18+2.5tT(t)=18+2.5tT(t)=18+2.5t, where ttt is time in minutes since the start of an experiment. The experiment runs from t=0t=0t=0 to t=40t=40t=40 minutes. What is the realistic domain for ttt?​

  1. t∈{0,1,2,…,40}t\in\{0,1,2,\dots,40\}t∈{0,1,2,…,40}
  2. t∈[0,40]t\in[0,40]t∈[0,40] (correct answer)
  3. t∈(0,40)t\in(0,40)t∈(0,40)
  4. t∈(−∞,40]t\in(-\infty,40]t∈(−∞,40]

Explanation: This question tests your understanding of how a function's domain relates both to its graph (which x-values have points) and to the real-world context (which values make sense practically). The domain is the set of allowed inputs, and for real-world functions, we distinguish: (1) mathematical domain (what the formula allows—like 'all reals' for polynomials), and (2) realistic domain (what makes sense in context—like 'positive integers' for number of items or 't ≥ 0' for time). The realistic domain is usually more restrictive because real-world constraints (can't have negative time, can't produce -5 items, can't work 1.7 hours if discrete) limit what mathematical values are meaningful. Discrete vs continuous domains depend on what you're measuring: if counting distinct objects (people, tickets, items sold), the domain is discrete—only integers work, graph as separate dots. If measuring continuous quantities (time, distance, temperature, money as a continuous quantity), the domain is continuous—any value in an interval works, graph as connected line/curve. The physical nature of the quantity determines which! You can have 2.7 hours but not 2.7 people. Time t in minutes is continuous, starting at 0 and ending at 40 as per the experiment's duration, including endpoints. Choice B correctly identifies the domain as t ∈ [0,40] based on the continuous measurement of time within the specified bounds. Choice A fails by suggesting discrete integers, but time isn't limited to whole minutes here. Domain determination framework: (1) Start with mathematical domain—what does the formula allow? (polynomials: all reals; √(expression): expression ≥ 0; 1/expression: expression ≠ 0; log(expression): expression > 0), (2) Apply context constraints—can variable be negative? Are there upper limits? Must it be integer?, (3) Combine all restrictions—domain is where ALL conditions are met. Example: f(x) = √(x - 3) for measuring length has mathematical domain x ≥ 3, and realistic domain also x ≥ 3 (lengths are non-negative, and formula requires x ≥ 3, so both agree). Discrete vs continuous quick-check: ask 'can there be in-between values?' If you're modeling number of students in a classroom, going from 20 to 21 students happens instantly (no 20.5 students exist)—discrete! If you're modeling temperature change from 20°C to 21°C, every value in between occurs—continuous! Context tells you: countable/distinct objects → discrete (dots on graph), measurable/varying quantities → continuous (line/curve on graph). This determines how you graph and what domain notation to use!

Question 5

If f(x)=(x+2)3f(x)=(x+2)^3f(x)=(x+2)3, solve f(x)=27f(x)=27f(x)=27 and express the solution as f−1(27)f^{-1}(27)f−1(27).​

  1. f−1(27)=1f^{-1}(27)=1f−1(27)=1 (correct answer)
  2. f−1(27)=5f^{-1}(27)=5f−1(27)=5
  3. f−1(27)=3f^{-1}(27)=3f−1(27)=3
  4. f−1(27)=−1f^{-1}(27)=-1f−1(27)=−1

Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. To solve f(x) = 27 where f(x) = (x+2)³, we need to find the x-value that makes (x+2)³ = 27. Taking the cube root of both sides gives x+2 = ∛27 = 3, so x = 3-2 = 1. This means f(1) = 27, which by definition of inverse means f⁻¹(27) = 1. Choice B correctly identifies f⁻¹(27) = 1 by solving the equation (x+2)³ = 27 to find x = 1. Choice A incorrectly gets x = 5 (perhaps from 3+2 instead of 3-2), while Choice C gives 3 without subtracting 2. The swap-and-solve recipe gives the general inverse: from y = (x+2)³, swap to x = (y+2)³, solve to get y = ∛x - 2, so f⁻¹(x) = ∛x - 2. Plugging in x = 27: f⁻¹(27) = ∛27 - 2 = 3 - 2 = 1 ✓. Inverse thinking: f adds 2 then cubes, so f⁻¹ must cube root then subtract 2!

Question 6

Convert the system to matrix form AX=BAX=BAX=B (use variable order x,y,zx, y, zx,y,z): [ \begin{cases} 2x+y-z=5\ x-3y+2z=1\ 4x+y+z=9 \end{cases} ]

  1. [2111−3241−1][xyz]=[519]\begin{bmatrix}2&1&1\\1&-3&2\\4&1&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\9\end{bmatrix}​214​1−31​12−1​​​xyz​​=​519​​
  2. [21−11−32411][xyz]=[519]\begin{bmatrix}2&1&-1\\1&-3&2\\4&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\9\end{bmatrix}​214​1−31​−121​​​xyz​​=​519​​ (correct answer)
  3. [21−11−32411][xzy]=[519]\begin{bmatrix}2&1&-1\\1&-3&2\\4&1&1\end{bmatrix}\begin{bmatrix}x\\z\\y\end{bmatrix}=\begin{bmatrix}5\\1\\9\end{bmatrix}​214​1−31​−121​​​xzy​​=​519​​
  4. [21−1−312411][xyz]=[519]\begin{bmatrix}2&1&-1\\-3&1&2\\4&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\9\end{bmatrix}​2−34​111​−121​​​xyz​​=​519​​

Explanation: This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX = B, where A is the coefficient matrix, X is the variable vector, and B is the constant vector. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, the system 2x + y - z = 5, x - 3y + 2z = 1, and 4x + y + z = 9 becomes [[2, 1, -1], [1, -3, 2], [4, 1, 1]] times [x, y, z] = [5, 1, 9]. Matrix multiplication recovers the original equations! To convert this system to matrix form: (1) Write coefficient matrix A by using equation 1 coefficients as row 1, equation 2 as row 2, equation 3 as row 3, with columns for each variable in order: A = [[2, 1, -1], [1, -3, 2], [4, 1, 1]] (first column x: 2, 1, 4; second y: 1, -3, 1; third z: -1, 2, 1). (2) Write variable vector X = [x, y, z] as column (order matches columns of A). (3) Write constant vector B = [5, 1, 9] as column (matching equation order). (4) Combine: [[2, 1, -1], [1, -3, 2], [4, 1, 1]] times [x, y, z] = [5, 1, 9]. This matrix equation is equivalent to the original system! Choice A correctly constructs the coefficient matrix A with proper row-column organization, variable vector X in correct order, and constant vector B matching the equations. Choice B has the z-coefficients in the third row wrong (should be +1, not -1). Remember: check signs carefully—z in third equation is +z, so +1 in A[3,3]. Don't overlook signs! Matrix equation construction recipe: (1) Label equations (first, second, third) and variables (x, y, z in consistent order), (2) Build matrix A: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write X as column vector with variables in same order as A's columns, (4) Write B as column vector with constants in same order as equations, (5) Combine as AX = B. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication AX mentally or on paper: the first row of A times X should give the left side of equation 1, and so on. If [[2, 1, -1], [1, -3, 2], [4, 1, 1]] times [x, y, z] gives [2x + y - z, x - 3y + 2z, 4x + y + z], and B = [5, 1, 9], then it matches. Matrix multiplication recovers the system—this verification prevents errors! Keep up the excellent work!

Question 7

A student says synthetic division is more efficient than long division when the divisor is of the form (x−c)(x-c)(x−c). Use synthetic division to divide P(x)=x3−2x2−5x+6P(x)=x^3-2x^2-5x+6P(x)=x3−2x2−5x+6 by (x−3).(x-3).(x−3). Set up with c=3c=3c=3 and coefficients 1,−2,−5,61,-2,-5,61,−2,−5,6. What are the quotient and remainder (remainder equals P(3)P(3)P(3))?

  1. Quotient x2+x−2x^2+x-2x2+x−2, remainder 000 (correct answer)
  2. Quotient x2−5x+10x^2-5x+10x2−5x+10, remainder 000
  3. Quotient x2+x−2x^2+x-2x2+x−2, remainder 666
  4. Quotient x2+x+2x^2+x+2x2+x+2, remainder 000

Explanation: This question tests your understanding of synthetic division—a streamlined shortcut for dividing polynomials by linear divisors of the form (x−c)(x - c)(x−c) that's much faster than polynomial long division. Synthetic division is an efficient algorithm for polynomial division when the divisor is (x−c)(x - c)(x−c): instead of the complex long division setup, you just write the value ccc (from x−cx - cx−c) on the left and the polynomial's coefficients on the right, then follow a simple multiply-and-add pattern that produces the quotient coefficients and remainder in one bottom row. To divide x3−2x2−5x+6x^3 - 2x^2 - 5x + 6x3−2x2−5x+6 by (x−3)(x - 3)(x−3) using synthetic division: identify c=3c = 3c=3, write coefficients 1,−2,−5,61, -2, -5, 61,−2,−5,6; bring down 111, multiply by 333 for 333, add to −2-2−2 for 111; multiply 111 by 333 for 333, add to −5-5−5 for −2-2−2; multiply −2-2−2 by 333 for −6-6−6, add to 666 for 000; quotient x2+x−2x^2 + x - 2x2+x−2 with remainder 000, and P(3)=27−18−15+6=0P(3) = 27 - 18 - 15 + 6 = 0P(3)=27−18−15+6=0 confirms. Choice A correctly executes the synthetic division algorithm and reads the quotient and remainder accurately. Choice C misreads the remainder as 666: the last bottom row value is the remainder, so verify with direct substitution if unsure. Synthetic division step-by-step: from divisor (x−c)(x - c)(x−c), identify ccc; write coefficients in descending order, including zeros if needed; bring down first, multiply by ccc, add to next, repeat; read quotient from bottom row except last, which is remainder. Awesome progress—zero remainder means you've found a factor, perfect for solving polynomials!

Question 8

A population of bacteria grows according to the model P(t)=500⋅20.3tP(t) = 500 \cdot 2^{0.3t}P(t)=500⋅20.3t, where ttt is time in hours. If the population reaches 8000 bacteria, which expression gives the exact time when this occurs?

  1. log⁡2(16)0.3\frac{\log_2(16)}{0.3}0.3log2​(16)​
  2. log⁡(16)0.3log⁡(2)\frac{\log(16)}{0.3 \log(2)}0.3log(2)log(16)​
  3. ln⁡(16)0.3ln⁡(2)\frac{\ln(16)}{0.3 \ln(2)}0.3ln(2)ln(16)​ (correct answer)
  4. log⁡2(8000)0.3\frac{\log_2(8000)}{0.3}0.3log2​(8000)​

Explanation: Setting up the equation: 500⋅20.3t=8000500 \cdot 2^{0.3t} = 8000500⋅20.3t=8000. Dividing both sides by 500: 20.3t=162^{0.3t} = 1620.3t=16. Taking the natural logarithm of both sides: ln⁡(20.3t)=ln⁡(16)\ln(2^{0.3t}) = \ln(16)ln(20.3t)=ln(16). Using the logarithm property: 0.3t⋅ln⁡(2)=ln⁡(16)0.3t \cdot \ln(2) = \ln(16)0.3t⋅ln(2)=ln(16). Solving for ttt: t=ln⁡(16)0.3ln⁡(2)t = \frac{\ln(16)}{0.3 \ln(2)}t=0.3ln(2)ln(16)​. Choice A uses the wrong logarithm base conversion. Choice B uses common log instead of natural log but has the correct structure. Choice D fails to simplify by dividing out the initial population.

Question 9

The expression x3−6x2+12x−8x^3 - 6x^2 + 12x - 8x3−6x2+12x−8 has a hidden structure. Which form reveals this structure most clearly?

  1. x2(x−6)+4(3x−2)x^2(x - 6) + 4(3x - 2)x2(x−6)+4(3x−2), showing factoring by grouping
  2. (x−2)3(x - 2)^3(x−2)3, showing it as a perfect cube (correct answer)
  3. (x−2)(x2−4x+4)(x - 2)(x^2 - 4x + 4)(x−2)(x2−4x+4), showing linear and quadratic factors
  4. x(x2−6x+12)−8x(x^2 - 6x + 12) - 8x(x2−6x+12)−8, showing a quadratic nested structure

Explanation: The expression x3−6x2+12x−8x^3 - 6x^2 + 12x - 8x3−6x2+12x−8 matches the pattern a3−3a2b+3ab2−b3=(a−b)3a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3a3−3a2b+3ab2−b3=(a−b)3 with a=xa = xa=x and b=2b = 2b=2. This gives (x−2)3(x - 2)^3(x−2)3, which can be verified by expansion. Choice A attempts factoring by grouping but doesn't work correctly. Choice C shows partial factoring but misses that x2−4x+4=(x−2)2x^2 - 4x + 4 = (x - 2)^2x2−4x+4=(x−2)2. Choice D factors out xxx but doesn't reveal the perfect cube structure and the remaining expression doesn't factor nicely.

Question 10

Factor and sketch g(x)=x3−4x2−x+4g(x)=x^3-4x^2-x+4g(x)=x3−4x2−x+4 showing all real zeros and the end behavior.

  1. g(x)=(x−1)(x−4)(x+1)g(x)=(x-1)(x-4)(x+1)g(x)=(x−1)(x−4)(x+1). Zeros: x=1,4,−1x=1,4,-1x=1,4,−1. End behavior: left down, right up. (correct answer)
  2. g(x)=(x+1)(x−4)(x−1)g(x)=(x+1)(x-4)(x-1)g(x)=(x+1)(x−4)(x−1). Zeros: x=−1,1,4x=-1,1,4x=−1,1,4. End behavior: left up, right down.
  3. g(x)=(x−1)2(x−4)g(x)=(x-1)^2(x-4)g(x)=(x−1)2(x−4). Zeros: x=1x=1x=1 (touch), x=4x=4x=4 (cross). End behavior: left down, right up.
  4. g(x)=(x+1)(x2−4)g(x)=(x+1)(x^2-4)g(x)=(x+1)(x2−4). Zeros: x=−1,−2,2x=-1,-2,2x=−1,−2,2. End behavior: left down, right up.

Explanation: This question tests your ability to graph polynomial functions by identifying zeros from factorizations and determining end behavior from the leading term's degree and coefficient. End behavior depends ONLY on the leading term ax^n, because for large |x|, this term dominates all others: in p(x) = 2x⁴ - 100x³ + 500x - 1000, for x = 1000, the 2x⁴ term equals 2 trillion while other terms are relatively tiny. Factoring g(x) = x^3 - 4x^2 - x + 4 gives (x-1)(x-4)(x+1), with zeros x = 1, 4, -1 (all cross), and degree 3 odd positive means left down, right up. Choice A correctly provides the factored form, zeros, and end behavior. Choice B fails by stating incorrect end behavior (left up, right down) despite the same factors, which would require a negative leading coefficient. End behavior shortcut: (1) find degree n—count highest power, (2) find sign of leading coefficient a—look at coefficient of x^n term, (3) apply pattern: even n = both ends match (up if a > 0, down if a < 0); odd n = ends opposite (if a > 0: ↓↑, if a < 0: ↑↓). The complete polynomial graphing checklist: (1) Find zeros: set each factor equal to zero (watch signs!), (2) Determine multiplicity: count factor appearances, note cross (odd) or touch (even) at each zero, (3) Find y-intercept: evaluate f(0), (4) Determine end behavior: degree + leading coefficient sign, (5) Plot zeros and y-intercept on axes, (6) Sketch smooth curve through/touching zeros with correct end behavior.

Question 11

In physics, the period of a simple pendulum can be approximated by T=2πLgT=2\pi\sqrt{\dfrac{L}{g}}T=2πgL​​, where TTT is the period, LLL is the pendulum length, and ggg is gravitational acceleration. Rearrange T=2πLgT=2\pi\sqrt{\dfrac{L}{g}}T=2πgL​​ to solve for LLL.

  1. L=gT22πL=\dfrac{gT^2}{2\pi}L=2πgT2​
  2. L=4π2gT2L=\dfrac{4\pi^2}{gT^2}L=gT24π2​
  3. L=gT24π2L=\dfrac{gT^2}{4\pi^2}L=4π2gT2​ (correct answer)
  4. L=gT4π2L=\dfrac{gT}{4\pi^2}L=4π2gT​

Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in A = πr²), you'll need square roots (r = √(A/π)). If it appears in a denominator (like f in 1/f = 1/a + 1/b), you'll need to clear fractions first. If it appears with different powers (like t in s = ut + (1/2)at²), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. Starting with T = 2π√(L/g), we need to isolate L. First divide both sides by 2π: T/(2π) = √(L/g). Then square both sides: T²/(4π²) = L/g. Finally, multiply both sides by g: gT²/(4π²) = L, which gives us L = gT²/(4π²). Choice B correctly isolates L through division, squaring, and multiplication to get L = gT²/(4π²). Choice A forgets to square T, Choice C inverts the entire expression, and Choice D has the wrong denominator (2π instead of 4π²). Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in xy + xz = w), factor it out first: x(y + z) = w, then x = w/(y + z). If you don't factor, you'll struggle to isolate! Also, when taking square roots of a variable, remember ± if the formula context allows both positive and negative (though often context restricts to positive only, like radius r ≥ 0). Physics and geometry formulas usually want positive values only!

Question 12

Rewrite the exponential equation 103=100010^3 = 1000103=1000 in logarithmic form.

  1. log⁡3(10)=1000\log_{3}(10)=1000log3​(10)=1000
  2. log⁡10(3)=1000\log_{10}(3)=1000log10​(3)=1000
  3. log⁡10(1000)=3\log_{10}(1000)=3log10​(1000)=3 (correct answer)
  4. log⁡1000(10)=3\log_{1000}(10)=3log1000​(10)=3

Explanation: This question tests your understanding that logarithms and exponentials are inverse operations—logarithms undo exponentiation and vice versa, just like square roots undo squaring. As inverse operations, logarithms and exponents cancel each other: log_b(b^x) = x for any x (the log undoes the exponent), and b^(log_b(x)) = x for x > 0 (the exponent undoes the log). These inverse properties are incredibly useful for simplification: log₃(3⁵) immediately simplifies to 5, and 7^(log₇(20)) immediately simplifies to 20. No calculation needed—they just undo each other! To rewrite 10^3 = 1000 in logarithmic form, identify the base 10, exponent 3, and result 1000, so it becomes log_{10}(1000) = 3. Choice B correctly converts to logarithmic form with the base as 10, argument as 1000, and value as 3. Choice A swaps the argument and value incorrectly, making it log_{10}(3) = 1000, which would imply 10^{1000} = 3, but that's not true. Converting between forms: identify the three parts—base, exponent, and result. In b^y = x: base is b, exponent is y, result is x. In log_b(x) = y: base is b (subscript), argument is x (inside the log), value is y (what log equals). The exponent in exponential form BECOMES the log value, and the result BECOMES the argument. Example: 5³ = 125 has base 5, exponent 3, result 125, so log₅(125) = 3 has base 5, argument 125, value 3. The positions shift but the numbers stay the same!

Question 13

A savings account has an annual growth factor of 1.061.061.06. Which is the approximate monthly growth factor? (Use 1.061/12≈1.00491.06^{1/12} \approx 1.00491.061/12≈1.0049.)

  1. 1.06/12≈0.08831.06/12 \approx 0.08831.06/12≈0.0883
  2. 1.061/12≈1.00491.06^{1/12} \approx 1.00491.061/12≈1.0049 (correct answer)
  3. 1.061/6≈1.00971.06^{1/6} \approx 1.00971.061/6≈1.0097
  4. 1.0612≈2.011.06^{12} \approx 2.011.0612≈2.01

Explanation: This question tests your ability to identify the correct monthly growth factor when given an annual growth factor, using the concept of compound interest and fractional exponents. The key insight is that to find a monthly factor that compounds to give the annual factor, we need the 12th root of the annual factor: if the monthly factor is m, then m^12 = 1.06, so m = (1.06)^(1/12). Using a calculator, (1.06)^(1/12) ≈ 1.0049, meaning about 0.49% monthly growth. This makes sense: 0.49% monthly compounded 12 times gives approximately 6% annually! To verify: (1) Monthly factor m must satisfy m^12 = 1.06 (compound monthly to get annual), (2) Taking 12th root of both sides: m = (1.06)^(1/12), (3) Calculate: (1.06)^(1/12) ≈ 1.0049, (4) Check: 1.0049^12 ≈ 1.0600 ✓, (5) So monthly growth is about 0.49% (since 1.0049 = 1 + 0.0049). The fractional exponent gives the exact compound rate! Choice C correctly identifies (1.06)^(1/12) ≈ 1.0049 as the monthly growth factor. Choice A incorrectly divides 1.06 by 12 to get 0.0883, but this would mean the account SHRINKS each month (factor less than 1)! This error comes from confusing growth factors with growth rates: the annual RATE is 6% = 0.06, but the annual FACTOR is 1.06 = 1 + 0.06. Even if we divided the rate 0.06/12 = 0.005, the monthly factor would be 1.005, not 0.0883! For compound interest conversions: (1) Annual factor to monthly: take 12th root, (2) Annual factor to daily: take 365th root, (3) General rule: (annual factor)^(1/n) gives the factor for 1/n of a year. Never divide the factor itself—always use fractional exponents for compound interest! The difference between simple and compound interest is crucial: simple interest would divide the rate, but compound interest requires the root!

Question 14

Find xxx in 15e2x=40015e^{2x}=40015e2x=400. Give the exact solution as a logarithm and an approximate value using technology.

  1. x=ln⁡(40015)2≈1.642x=\dfrac{\ln\left(\frac{400}{15}\right)}{2}\approx 1.642x=2ln(15400​)​≈1.642 (correct answer)
  2. x=ln⁡(40015)≈3.284x=\ln\left(\frac{400}{15}\right)\approx 3.284x=ln(15400​)≈3.284
  3. x=ln⁡(15400)2≈−1.642x=\dfrac{\ln\left(\frac{15}{400}\right)}{2}\approx -1.642x=2ln(40015​)​≈−1.642
  4. x=log⁡(40015)2≈0.713x=\dfrac{\log\left(\frac{400}{15}\right)}{2}\approx 0.713x=2log(15400​)​≈0.713

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 15e^(2x) = 400 where e is the base and the coefficient of x is 2, natural logarithms are the perfect tool: first isolate the exponential by dividing both sides by 15 to get e^(2x) = 400/15 ≈ 26.667, then take ln of both sides to get ln(e^(2x)) = ln(400/15), and using the inverse property ln(e^(2x)) = 2x, we get 2x = ln(400/15), so x = ln(400/15)/2. To solve 15e^(2x) = 400 step by step: divide by 15 to get e^(2x) = 26.667, take natural log of both sides: ln(e^(2x)) = ln(26.667), apply the inverse property to get 2x = ln(26.667) ≈ 3.283, and divide by 2: x = ln(26.667)/2 = ln(400/15)/2 ≈ 3.283/2 ≈ 1.642. Choice A correctly isolates the exponential, takes the natural logarithm, and divides by 2 to get x = ln(400/15)/2 ≈ 1.642. Choice B forgets to divide by 2, giving x = ln(400/15) ≈ 3.284, which would mean e^(2x) = e^6.568, far larger than 26.667. Isolation before logarithms: ALWAYS isolate the exponential expression e^(2x) before taking logarithms. If you have 15e^(2x) = 400, first divide by 15 to get e^(2x) = 26.667, THEN take ln. When working with base e, always use natural log (ln) for the cleanest solution—the inverse property ln(e^x) = x makes the algebra straightforward!

Question 15

A formula for the area of a shape is A=π (r+2)2.A = \pi\,(r+2)^2.A=π(r+2)2. If you chunk (r+2)(r+2)(r+2) as a single entity, which interpretation best describes the structure and what depends on rrr?

  1. AAA is the product of π\piπ and the chunk (r+2)2(r+2)^2(r+2)2; π\piπ does not depend on rrr, while (r+2)2(r+2)^2(r+2)2 does. (correct answer)
  2. AAA is the sum of π\piπ and (r+2)2(r+2)^2(r+2)2, so increasing rrr only changes the π\piπ part.
  3. AAA shows that only the +2 depends on rrr, because it is closest to rrr.
  4. AAA means rrr is squared first and then 2 is added, so the chunk should be r2+2r^2+2r2+2.

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 get complex, chunking (viewing sub-expressions as single units) reveals structure: for A = π(r+2)^2, viewing (r+2) as a single chunk u gives A = πu^2, clearly showing it's π times a squared quantity. In the expression A = π(r+2)^2, we have a product of two factors: (1) The constant π—completely independent of r, (2) The chunk (r+2)^2—this entire expression depends on r because r appears inside the parentheses that are being squared. The structure is multiplicative: A = π × (r+2)^2, where π scales the squared term. Choice A correctly interprets this as a product where π doesn't depend on r while (r+2)^2 does—this reveals the geometric meaning: area equals π times the square of the adjusted radius (r+2). Choice B misreads this as addition π + (r+2)^2, but there's no plus sign—when π is written next to parentheses, it means multiplication, and Choice D misunderstands order of operations, claiming we square r first then add 2, but parentheses mean we add 2 to r first, then square the entire sum. Chunking strategy for geometric formulas: (1) Constants like π, 2π, or 4π often represent geometric factors that don't depend on the variable dimensions, (2) Squared or cubed chunks like (expression)^2 represent area or volume scaling, (3) Variables inside parentheses affect the entire chunk—here, changing r shifts the entire squared quantity. Understanding structure reveals behavior: since A = π(r+2)^2, the area grows quadratically as r increases—not because of r^2, but because (r+2)^2 = r^2+4r+4, where all terms contribute to growth!

Question 16

Which equation is equivalent to e2ln⁡(x)−ln⁡(3)=12e^{2\ln(x)-\ln(3)} = 12e2ln(x)−ln(3)=12 where x>0x > 0x>0?

  1. ln⁡(x2)−ln⁡(3)=12\ln(x^2) - \ln(3) = 12ln(x2)−ln(3)=12
  2. x2−3=12x^2 - 3 = 12x2−3=12
  3. 2x−3=ln⁡(12)2x - 3 = \ln(12)2x−3=ln(12)
  4. x23=12\frac{x^2}{3} = 123x2​=12 (correct answer)

Explanation: When you encounter equations involving both exponentials and logarithms, the key strategy is to use the properties of logarithms to simplify the exponent before dealing with the exponential equation. Starting with e2ln⁡(x)−ln⁡(3)=12e^{2\ln(x)-\ln(3)} = 12e2ln(x)−ln(3)=12, first simplify the exponent using logarithm properties. Since 2ln⁡(x)=ln⁡(x2)2\ln(x) = \ln(x^2)2ln(x)=ln(x2) and ln⁡(a)−ln⁡(b)=ln⁡(ab)\ln(a) - \ln(b) = \ln(\frac{a}{b})ln(a)−ln(b)=ln(ba​), you can rewrite the exponent as: 2ln⁡(x)−ln⁡(3)=ln⁡(x2)−ln⁡(3)=ln⁡(x23)2\ln(x) - \ln(3) = \ln(x^2) - \ln(3) = \ln(\frac{x^2}{3})2ln(x)−ln(3)=ln(x2)−ln(3)=ln(3x2​) This transforms the equation to eln⁡(x23)=12e^{\ln(\frac{x^2}{3})} = 12eln(3x2​)=12. Since eln⁡(y)=ye^{\ln(y)} = yeln(y)=y for any positive yyy, this simplifies to x23=12\frac{x^2}{3} = 123x2​=12, which is answer choice D. Let's examine why the other options are incorrect. Choice A, ln⁡(x2)−ln⁡(3)=12\ln(x^2) - \ln(3) = 12ln(x2)−ln(3)=12, represents only the exponent from the original equation, not the complete simplified form after applying the exponential. Choice B, x2−3=12x^2 - 3 = 12x2−3=12, incorrectly converts the logarithmic subtraction ln⁡(x2)−ln⁡(3)\ln(x^2) - \ln(3)ln(x2)−ln(3) into algebraic subtraction x2−3x^2 - 3x2−3. Choice C, 2x−3=ln⁡(12)2x - 3 = \ln(12)2x−3=ln(12), makes the error of treating 2ln⁡(x)2\ln(x)2ln(x) as 2x2x2x instead of ln⁡(x2)\ln(x^2)ln(x2), completely mishandling the logarithm properties. Remember this pattern: when you see elogarithmic expression=numbere^{\text{logarithmic expression}} = \text{number}elogarithmic expression=number, always simplify the logarithmic expression first using properties like aln⁡(x)=ln⁡(xa)a\ln(x) = \ln(x^a)aln(x)=ln(xa) and ln⁡(a)−ln⁡(b)=ln⁡(ab)\ln(a) - \ln(b) = \ln(\frac{a}{b})ln(a)−ln(b)=ln(ba​), then use the fact that eee and ln⁡\lnln cancel each other.

Question 17

A geometric sequence is given: 3,6,12,24,…3, 6, 12, 24, \dots3,6,12,24,… Let g(n)g(n)g(n) give the nnnth term with g(1)=3g(1)=3g(1)=3. Which exponential function matches this sequence?

  1. g(n)=3⋅2n−1g(n)=3\cdot 2^{n-1}g(n)=3⋅2n−1 (correct answer)
  2. g(n)=3⋅3n−1g(n)=3\cdot 3^{n-1}g(n)=3⋅3n−1
  3. g(n)=6⋅2ng(n)=6\cdot 2^ng(n)=6⋅2n
  4. g(n)=2n+1g(n)=2n+1g(n)=2n+1

Explanation: This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. From a table, first determine which type: calculate differences between consecutive y-values (if constant → linear with slope = that difference), and calculate ratios (if constant → exponential with base = that ratio). This identification step is crucial—you can't construct the right function if you don't know which type it is! Once identified, extract the parameters (slope and intercept for linear, initial value and base for exponential) and write the formula. For the geometric sequence with constant ratio 2 and g(1)=3, adjust to g(n)=3·2^{n-1} to fit the exponent starting at n=1. Choice A correctly constructs the exponential function with a=3 and b=2 adjusted for n-1 from the sequence. A distractor like choice D uses b=3, but 3·3^{1-1}=3, then 3·3^1=9≠6—check the common ratio matches b. Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = y_{x+1}/y_x—should be constant for exponential, (3) Write f(x) = a·b^x, (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100·(1.1)^x. The ratio test both identifies the type and gives you the base!

Question 18

For a quadratic ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0 with real coefficients, the discriminant b2−4acb^2-4acb2−4ac determines whether the two complex solutions (counting multiplicity) are 2 distinct real, 1 repeated real, or 2 complex conjugates. If a quadratic has discriminant b2−4ac<0b^2-4ac<0b2−4ac<0, which must be true?

  1. It has exactly one complex solution (since the discriminant is negative).
  2. It has exactly two complex solutions that are conjugates of each other. (correct answer)
  3. It has exactly two real solutions.
  4. It has no solutions in the complex numbers.

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. If discriminant <0 for real-coefficient quadratic, there are two complex conjugate solutions (not real, but still exactly two). Choice B correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies conjugates for negative discriminant. Choice A says exactly one complex solution since discriminant negative—this ignores that for real coefficients, complex roots come in pairs; can't have odd number. To apply this transferable strategy: (1) Negative disc means conjugates p ± qi. (2) Real coefficients ensure pair. (3) Total 2. Example: if disc=-16, q=sqrt(16)/(2a)=2/|a| or similar. You're excelling—believe in yourself!

Question 19

Factor the quadratic function f(x)=x2+7x+12f(x)=x^2+7x+12f(x)=x2+7x+12 and use the zero product property to find its zeros. Then identify the x-intercepts of the graph.

  1. f(x)=(x+2)(x+6)f(x)=(x+2)(x+6)f(x)=(x+2)(x+6); zeros: x=−2,−6x=-2,-6x=−2,−6; x-intercepts: (−2,0),(−6,0)(-2,0),(-6,0)(−2,0),(−6,0)
  2. f(x)=(x+3)(x−4)f(x)=(x+3)(x-4)f(x)=(x+3)(x−4); zeros: x=3,−4x=3,-4x=3,−4; x-intercepts: (3,0),(−4,0)(3,0),(-4,0)(3,0),(−4,0)
  3. f(x)=(x+3)(x+4)f(x)=(x+3)(x+4)f(x)=(x+3)(x+4); zeros: x=−3,−4x=-3,-4x=−3,−4; x-intercepts: (−3,0),(−4,0)(-3,0),(-4,0)(−3,0),(−4,0) (correct answer)
  4. f(x)=(x−3)(x−4)f(x)=(x-3)(x-4)f(x)=(x−3)(x−4); zeros: x=3,4x=3,4x=3,4; x-intercepts: (3,0),(4,0)(3,0),(4,0)(3,0),(4,0)

Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s. These are the x-intercepts of the parabola—points (r, 0) and (s, 0) where the graph crosses the x-axis. Factoring transforms the quadratic from a form where zeros are hidden (standard form) to a form where they're obvious (factored form)! To factor x squared + 7x + 12 and find zeros: (1) Look for two numbers that multiply to 12 (constant term) and add to 7 (middle coefficient): factors of 12 are 1 and 12 (sum 13, no), 2 and 6 (sum 8, no), 3 and 4 (sum 7, yes!). (2) Write factored form: (x + 3)(x + 4). (3) Find zeros using zero product property: set (x + 3)(x + 4) = 0, so x + 3 = 0 giving x = -3, or x + 4 = 0 giving x = -4. The zeros are x = -3 and x = -4. (4) X-intercepts are (-3, 0) and (-4, 0). The factored form makes zeros immediate—no quadratic formula needed! Choice A correctly factors the quadratic and identifies both zeros by setting each factor equal to zero and solving. Choice B has a sign error in the zeros: from factor (x + 3), the zero is x = -3 (not x = 3). The sign flips! From (x - r), zero is x = r, so the sign in the factor is OPPOSITE to the zero. Think: what value makes (x + 3) equal zero? x = -3. This sign relationship trips everyone up initially—practice makes it automatic! Factoring trinomial x squared + bx + c: find two numbers that (1) multiply to c (constant term), (2) add to b (middle coefficient). Those numbers go in factors: (x + first number)(x + second number). Example: x squared - 5x + 6, find factors of 6 that add to -5: that's -2 and -3 (multiply to 6, add to -5). Factored: (x - 2)(x - 3). Zeros: x = 2, 3. For leading coefficient not 1 (like 2x squared + 7x + 3), use grouping or trial combinations. From factored form to zeros: the sign trick is crucial. (x - r) gives zero x = r (opposite sign), (x + s) = (x - (-s)) gives zero x = -s (opposite sign). From (x - 5), zero is x = 5. From (x + 2), zero is x = -2. Set each factor equal to zero: (x - 5) = 0 means x = 5, (x + 2) = 0 means x = -2. The zeros always have opposite sign from what appears in the factored form (unless the factor is written differently). Master this sign relationship!

Question 20

A parabola and a line intersect at points AAA and BBB. If the parabola is y=x2+2x−3y = x^2 + 2x - 3y=x2+2x−3 and the midpoint of segment ABABAB is (1,2)(1, 2)(1,2), what is the equation of the line?

  1. y=4x−2y = 4x - 2y=4x−2 (correct answer)
  2. y=3x−1y = 3x - 1y=3x−1
  3. y=2x+0y = 2x + 0y=2x+0
  4. y=5x−3y = 5x - 3y=5x−3

Explanation: Let the line be y = mx + b, and let the intersection points be (x₁, y₁) and (x₂, y₂). Since the midpoint is (1, 2), we have (x₁ + x₂)/2 = 1, so x₁ + x₂ = 2. Setting x² + 2x - 3 = mx + b gives x² + (2 - m)x + (-3 - b) = 0. By Vieta's formulas, x₁ + x₂ = -(2 - m) = m - 2. Since x₁ + x₂ = 2, we have m - 2 = 2, so m = 4. For the y-coordinate of the midpoint: (y₁ + y₂)/2 = 2, so y₁ + y₂ = 4. Since both points lie on the line y = 4x + b, we have y₁ = 4x₁ + b and y₂ = 4x₂ + b. Thus y₁ + y₂ = 4(x₁ + x₂) + 2b = 4(2) + 2b = 8 + 2b. Setting this equal to 4: 8 + 2b = 4, so 2b = -4, giving b = -2. Therefore, the line is y = 4x - 2. Choice B gives slope 3, not 4. Choice C gives slope 2, not 4. Choice D has the correct slope but wrong y-intercept.

Question 21

Use zeros to construct a rough graph of p(x)=−x(x+2)2(x−5).p(x)=-x(x+2)^2(x-5).p(x)=−x(x+2)2(x−5). Identify the zeros with multiplicities, determine crossing vs touching, and state the end behavior.

  1. Zeros: x=0x=0x=0 (mult. 1), x=−2x=-2x=−2 (mult. 2), x=5x=5x=5 (mult. 1). Crosses at 000 and 555, touches at −2-2−2. End behavior: as x→−∞x\to -\inftyx→−∞, p(x)→−∞p(x)\to -\inftyp(x)→−∞ and as x→+∞x\to +\inftyx→+∞, p(x)→−∞p(x)\to -\inftyp(x)→−∞. (correct answer)
  2. Zeros: x=0x=0x=0 (mult. 1), x=−2x=-2x=−2 (mult. 2), x=5x=5x=5 (mult. 1). Crosses at all three zeros. End behavior: both ends down.
  3. Zeros: x=0x=0x=0 (mult. 2), x=−2x=-2x=−2 (mult. 1), x=5x=5x=5 (mult. 1). Touches at 000, crosses at −2-2−2 and 555. End behavior: both ends down.
  4. Zeros: x=0x=0x=0 (mult. 1), x=−2x=-2x=−2 (mult. 2), x=5x=5x=5 (mult. 1). Crosses at 000 and 555, touches at −2-2−2. End behavior: left up, right down.

Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = -x(x + 2)^2(x - 5), zeros are x = 0 (multiplicity 1, crosses), x = -2 (multiplicity 2, touches), x = 5 (multiplicity 1, crosses), degree 4 even negative, both ends down. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B assumes crossing at all but misses that even multiplicity touches—count exponents carefully to avoid this! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 22

The temperature in a greenhouse follows the function T(h)=68+12sin⁡(πh12)T(h) = 68 + 12\sin\left(\frac{\pi h}{12}\right)T(h)=68+12sin(12πh​), where TTT is the temperature in degrees Fahrenheit and hhh is the number of hours after midnight. Based on this model, which feature correctly describes the temperature pattern?

  1. The temperature oscillates between 56°F and 80°F with a period of 12 hours, reaching maximum at 6 AM.
  2. The temperature oscillates between 56°F and 80°F with a period of 24 hours, reaching maximum at 6 AM. (correct answer)
  3. The temperature oscillates between 68°F and 80°F with a period of 24 hours, reaching maximum at noon.
  4. The temperature oscillates between 60°F and 76°F with a period of 12 hours, reaching maximum at midnight.

Explanation: The function has amplitude 12, so temperature ranges from 68−12=56°F68-12=56°F68−12=56°F to 68+12=80°F68+12=80°F68+12=80°F. The period is 2ππ/12=24\frac{2\pi}{\pi/12} = 24π/122π​=24 hours. The maximum occurs when sin⁡(πh12)=1\sin\left(\frac{\pi h}{12}\right) = 1sin(12πh​)=1, which happens when πh12=π2\frac{\pi h}{12} = \frac{\pi}{2}12πh​=2π​, so h=6h = 6h=6 (6 AM). Choice A has the wrong period. Choice C has the wrong temperature range and maximum time. Choice D has wrong temperature range, period, and maximum time.

Question 23

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)!

Question 24

Consider the rational expression R(x)=x2+3x−2x−1.R(x)=\frac{x^2+3x-2}{x-1}.R(x)=x−1x2+3x−2​. If you view the numerator as a single entity, which statement correctly describes what happens to R(x)R(x)R(x) as xxx approaches 111 (without fully simplifying)?

  1. As x→1x\to 1x→1, the denominator approaches 000 while the numerator approaches a nonzero value, so R(x)R(x)R(x) becomes unbounded (blows up). (correct answer)
  2. As x→1x\to 1x→1, both numerator and denominator approach 000, so R(x)R(x)R(x) must approach 000.
  3. As x→1x\to 1x→1, the numerator approaches 000 and the denominator approaches 111, so R(x)R(x)R(x) approaches 000.
  4. As x→1x\to 1x→1, the denominator approaches 000 but the numerator is independent of xxx, so R(x)R(x)R(x) stays constant.

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 get complex, chunking (viewing sub-expressions as single units) reveals structure: for rational expressions like R(x) = (x² + 3x - 2)/(x - 1), viewing numerator and denominator as separate entities helps analyze behavior. In the expression R(x) = (x² + 3x - 2)/(x - 1), let's view the numerator as a single entity N(x) = x² + 3x - 2. As x approaches 1: (1) The denominator (x - 1) approaches 0, (2) The numerator N(1) = 1² + 3(1) - 2 = 1 + 3 - 2 = 2, which is nonzero, (3) We have a nonzero value divided by a number approaching zero, which makes R(x) grow without bound (approach ±∞). Choice A correctly identifies that as x → 1, the denominator approaches 0 while the numerator approaches a nonzero value (2), so R(x) becomes unbounded—this is a vertical asymptote! Choice B incorrectly claims both numerator and denominator approach 0, but we calculated N(1) = 2 ≠ 0—only the denominator goes to zero. Analyzing rational expressions by chunking: (1) View numerator and denominator as separate entities, (2) Evaluate each at the point of interest, (3) If denominator → 0 and numerator → nonzero, the function blows up (vertical asymptote), (4) If both → 0, you might have a removable discontinuity (requires further analysis). This chunking approach quickly identifies asymptotic behavior without full algebraic manipulation!

Question 25

A social media post's views follow V(d)=1500(2.4)dV(d) = 1500(2.4)^dV(d)=1500(2.4)d, where ddd represents days after posting. If the growth rate decreases so the new base becomes 1.81.81.8, by what factor will the views after 3 days be reduced compared to the original model?

  1. The views will be reduced by a factor of approximately 4.22
  2. The views will be reduced by a factor of approximately 1.33
  3. The views will be reduced by a factor of approximately 0.75
  4. The views will be reduced by a factor of approximately 2.37 (correct answer)

Explanation: When you encounter exponential function problems involving changes to the base, you're analyzing how different growth rates affect outcomes. The key is comparing function values at the same input. First, calculate the views after 3 days using the original model: V(3)=1500(2.4)3=1500(13.824)=20,736V(3) = 1500(2.4)^3 = 1500(13.824) = 20,736V(3)=1500(2.4)3=1500(13.824)=20,736 views. Next, find the views using the new model with base 1.8: Vnew(3)=1500(1.8)3=1500(5.832)=8,748V_{new}(3) = 1500(1.8)^3 = 1500(5.832) = 8,748Vnew​(3)=1500(1.8)3=1500(5.832)=8,748 views. To find the reduction factor, divide the original views by the new views: 20,7368,748≈2.37\frac{20,736}{8,748} \approx 2.378,74820,736​≈2.37. This means the views are reduced by a factor of approximately 2.37. Looking at the wrong answers: Choice A (4.22) likely comes from incorrectly dividing (2.4)3(2.4)^3(2.4)3 by (1.8)3(1.8)^3(1.8)3 without proper calculation: 13.8245.832≠4.22\frac{13.824}{5.832} ≠ 4.225.83213.824​=4.22. Choice B (1.33) might result from dividing the bases directly: 2.41.8=1.33\frac{2.4}{1.8} = 1.331.82.4​=1.33, but this ignores the exponent entirely. Choice C (0.75) represents the reciprocal of choice B, suggesting confusion about which direction the reduction goes. Remember: when comparing exponential functions with different bases, always evaluate both functions at the specific input value given, then compare the outputs. Don't just compare the bases themselves, as the exponent amplifies the difference significantly.