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Week 16 · Final study guide

Final Exam — Study Guide

College Algebra Free sample edition

Coverage — what the final covers

The final is cumulative over Weeks 1–15 and draws on all eight course objectives. It opens in Week 16, the finals week, which has no new module, and it closes at the time your instructor sets. There is no time limit, and you get one attempt. The questions appear one at a time, and you cannot go back once you move on.

Format: 40 questions, and every question counts the same. Thirty-four ask you to pick one option, four ask you to pick every option that applies, and two ask you to judge a statement true or false. Scratch paper and a calculator help, and one with ln and log keys covers the growth and decay questions.

What is in scope, with the number of final questions on each course objective:

Final coverage by course objective: the weeks that teach it, its main skills, and its number of questions out of 40
Course objective Weeks Main skills Questions
Real numbers, exponents, and expressions 1, used again in 12 The order of operations; the exponent rules, now with fraction exponents too 1
Linear equations and inequalities 2, used again in 5 and 15 Solving and checking by substitution; the flip rule; interval notation 1
Functions 3, used again in 9, 10, and 13 Evaluating at numeric and algebraic inputs; domain and range; composition 2
Lines and systems 4–5 Slope and intercepts; the forms of a line; solving a system 1
Polynomials and factoring 6, used again in 7, 10, and 11 Multiplying; the special products; factoring, greatest common factor first 1
Quadratic equations and functions 7 and 9, used again in 10 Solving four ways; the discriminant; the vertex, intercepts, and maximum or minimum 4
Polynomial, rational, and radical functions and expressions 10–12, used again in 14 End behavior; zeros and multiplicity; asymptotes and holes; rational and radical expressions and equations; extraneous solutions 15
Exponential and logarithmic functions 13–15 Growth and decay; compound interest; the log properties; exponential and log equations; doubling time and half-life 15

That is 30 of 40 questions on Weeks 10–15, split evenly between the rational-and-radical weeks and the exponential-and-logarithm weeks. Put most of your review there, then on the Week 9 parabola. The questions were drawn so that every objective appears at least once. The first-half skills also come back inside later material, such as the exponent rules with fraction exponents and checking a log solution by substitution.

Topic-by-topic review

Each topic gives the key idea in plain words, the course's own definitions, course examples to rehearse, and the mistakes students make most often with the cure for each. The deeper review for every topic is that week's chapter, lecture page, slide deck, and readings. Weeks 10–15 carry the weight of the exam, so they get the most space.

Weeks 1–6 — the first-half toolkit

These five objectives share six questions, and their rules run underneath everything after Week 7. Rehearse one core move per week.

  • Week 1, simplify. The order of operations is grouping symbols first, then exponents, then multiplication and division from left to right, then addition and subtraction from left to right. A negated power applies the exponent to the number first and the minus sign afterward. The exponent rules: product rule aᵐ · aⁿ = aᵐ⁺ⁿ, quotient rule aᵐ ⁄ aⁿ = aᵐ⁻ⁿ, power rule (aᵐ)ⁿ = aᵐⁿ, zero exponent a⁰ = 1, and negative exponent a⁻ⁿ = 1 ⁄ aⁿ. Rehearse: 2 + 3 · 4² = 2 + 48 = 50, and −4² = −16 but (−4)² = 16. Trap → cure: reading 2⁻² as a negative number. A negative exponent calls for a reciprocal, so 2⁻² = 1 ⁄ 4.
  • Week 2, solve. A solution is a value of the variable that makes a statement true. The flip rule: when both sides of an inequality are multiplied or divided by a negative number, the direction of the sign reverses. Rehearse: 3(x − 2) = 2x + 5 gives x = 11, and both sides check at 27. Trap → cure: solving −3x > 9 as x > −3. Circle the divisor; it is negative, so flip: x < −3.
  • Week 3, evaluate and compose. A function is a rule that pairs each input with exactly one output. The domain is every input that gives a real-number output. Composition (f ∘ g)(x) = f(g(x)) applies the inner function g first. Rehearse: for f(x) = 3x² − 2x + 5, f(−2) = 21, and the domain of √(x − 5) is [5, ∞). Trap → cure: evaluating 3(−2)² as 3(−4). The square of −2 is +4.
  • Weeks 4–5, lines and systems. Slope is rise over run, m = (y₂ − y₁) ⁄ (x₂ − x₁), with the y-values on top. In y = mx + b, m is the slope and b is the y-intercept. Perpendicular slopes multiply to −1. A solution of a system is an ordered pair that makes every equation true at once. Rehearse: 2x + 3y = 12 has intercepts (0, 4) and (6, 0), and x + y = 10 with x − y = 4 gives (7, 3). Trap → cure: swapping m and b. The slope multiplies x; the y-intercept stands alone.
  • Week 6, multiply and factor. Take out the greatest common factor first. The special products are (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², and (a + b)(a − b) = a² − b². A sum of squares such as x² + 16 is prime over the real numbers. Rehearse: (x − 1)(x − 5) = x² − 6x + 5, and 9x² − 25 = (3x + 5)(3x − 5). Trap → cure: writing (a + b)² as a² + b². The middle term 2ab is always there.

Weeks 7 and 9 — Quadratic equations and functions

  • Key idea: a quadratic equation asks where a parabola meets the x-axis, and the parabola itself answers the questions "how high?" and "when?" at its vertex.
  • Solving (Week 7): the zero-product property says that if a product of factors equals 0, at least one factor equals 0. The quadratic formula is x = (−b ± √(b² − 4ac)) ⁄ (2a). The discriminant b² − 4ac is positive for two real solutions, 0 for one repeated solution, and negative for no real solution.
  • Graphs (Week 9): a quadratic function can be written f(x) = ax² + bx + c with a not equal to zero; its graph is a parabola. The vertex (h, k) is the turning point. Vertex form is f(x) = a(x − h)² + k, with h taken from inside the parentheses with its sign reversed. From standard form, the vertex formula gives h = −b ⁄ (2a), and then k = f(h). When a is positive the parabola opens up and k is the minimum value; when a is negative it opens down and k is the maximum value. The y-intercept is (0, c).
  • Course examples to rehearse: x² − 6x + 5 = 0 gives x = 1 or x = 5 by factoring, by completing the square, and by the formula. As a function, x² − 6x + 5 has vertex (3, −4), minimum −4, y-intercept (0, 5), and vertex form (x − 3)² − 4. The function −x² + 4x + 1 has vertex (2, 5), a maximum. The launched object h(t) = −16t² + 64t + 80 peaks at t = 2 seconds at 144 feet.
  • Predictable mistakes → cures: (1) reading (x + 3)² − 5 as vertex (3, −5). Cure: write it as (x − (−3))² − 5, so the vertex is (−3, −5). (2) calling a downward parabola's vertex a minimum. Cure: read the sign of a first. (3) dividing both sides by x and losing x = 0. Cure: factor out the greatest common factor instead. (4) writing −b as −6 when b = −6. Cure: −(−6) = 6.

Week 10 — Polynomial and rational functions

  • Key idea: before you plot a single point, the formula tells you where the graph's ends go, where it meets the x-axis, and which lines it approaches.
  • Definitions: the leading term is the term with the highest power of the variable. End behavior is the direction the graph heads at its far left and far right, and for a polynomial it is decided by the leading term alone. An even degree sends both ends the same way; an odd degree sends them opposite ways; the sign of the leading coefficient decides which. The multiplicity of a zero is the exponent on its factor. At an odd multiplicity the graph crosses the x-axis; at an even multiplicity it touches and turns back.
  • Rational functions: factor first. A vertical asymptote comes from a zero of the denominator that is still there after common factors are cancelled. A hole is a single missing point where a factor cancelled. The horizontal asymptote comes from comparing degrees: numerator smaller gives y = 0, equal degrees give the ratio of the leading coefficients, and numerator larger gives none.
  • Course examples to rehearse: −2x⁴ + 3x² − 1 falls on both ends, and x³ − 5x + 2 falls on the left and rises on the right. (x − 1)²(x + 4) touches at x = 1 and crosses at x = −4. R(x) = (x + 1) ⁄ (x² − 4) has vertical asymptotes x = −2 and x = 2, horizontal asymptote y = 0, and intercepts (−1, 0) and (0, −1⁄4). (x² − x) ⁄ (x − 1) has a hole at (1, 1). (3x + 1) ⁄ (x − 5) has horizontal asymptote y = 3, and (x² + 1) ⁄ (x − 1) has none.
  • Predictable mistakes → cures: (1) reading end behavior from a lower-degree term. Cure: find the highest power and ignore the rest. (2) saying the graph "crosses" at an even-multiplicity zero. Cure: check the exponent on the factor first. (3) calling a cancelled factor's zero an asymptote. Cure: a cancelled factor leaves a hole. (4) offering y = 1 or a slanted line when the numerator's degree is larger. Cure: larger on top means no horizontal asymptote.

Week 11 — Rational expressions and equations

  • Key idea: a fraction with polynomials works like a number fraction, with one extra duty: keep track of the inputs that would make a denominator 0.
  • Definitions: a rational expression is one polynomial divided by another. An excluded value is an input that makes a denominator equal 0; it is found before any simplifying or solving and can never be a solution. A restriction such as x ≠ −3 travels with the simplified form. Only a common factor may be cancelled, never a term that is added or subtracted. An extraneous solution is a candidate produced by correct steps that still fails the original equation. A work rate is the fraction of a job done per unit of time, and rates add.
  • Course examples to rehearse: (x² − 9) ⁄ (x + 3) = x − 3 for x ≠ −3. The week's quotient (x² − 6x + 5) ⁄ (x − 1) = x − 5 for x ≠ 1, a hole at (1, −4), not a zero. 1 ⁄ x + 1 ⁄ (x + 1) = (2x + 1) ⁄ (x(x + 1)). x ⁄ (x − 2) = 2 ⁄ (x − 2) has no solution, because its only candidate is excluded. Pipes that fill a tank in 3 hours and 6 hours fill it together in 2 hours.
  • Predictable mistakes → cures: (1) "cancelling" the x in (x + 3) ⁄ x to get 3. Cure: x is a term, not a factor; at x = 1 the expression is 4. (2) multiplying the denominators instead of building the LCD from factors. Cure: factor every denominator first. (3) subtracting only the first term of the second numerator. Cure: put the whole numerator in parentheses before the minus. (4) adding times, 3 + 6 = 9 hours. Cure: add the rates; together must be faster than either alone.

Week 12 — Radicals, rational exponents, and radical equations

  • Key idea: a root undoes a power, and a fraction exponent is a root in disguise.
  • Definitions: the nth root ⁿ√a is the number whose nth power is a; the small n is the index. The product rule for radicals is √(ab) = √a · √b for a and b that are 0 or more. Simplified radical form leaves no perfect-square factor other than 1 under the root. A rational exponent names a root and a power: a^(m⁄n) is the nth root of a, raised to the power m. A radical equation is solved by isolating the radical, raising both sides to the index, solving, and checking every candidate.
  • Course examples to rehearse: √50 = 5√2 ≈ 7.071, and √48 = 4√3. 8^(2⁄3) = 4, 16^(3⁄4) = 8, and 8^(−2⁄3) = 1 ⁄ 4. x^(1⁄2) · x^(1⁄3) = x^(5⁄6). √(2x − 1) = 3 gives x = 5. √(x + 3) = x + 1 gives x = 1 only, because x = −2 makes the left side 1 and the right side −1.
  • Predictable mistakes → cures: (1) √(9 + 16) = 7. Cure: the radical bar groups, so add first: √25 = 5. (2) leaving √48 as 2√12. Cure: pull out the largest perfect square, 16. (3) cubing first in 16^(3⁄4). Cure: take the root first and keep the numbers small. (4) keeping both candidates after squaring. Cure: test each one in the original, because a principal square root is never negative.

Week 13 — Exponential functions

  • Key idea: a line adds the same amount at each step, and an exponential multiplies by the same factor at each step.
  • Definitions: an exponential function has the form f(x) = a · bˣ, with a ≠ 0, b > 0, and b ≠ 1. The initial value a is the output at x = 0. The growth factor b is the number each output is multiplied by when x goes up by 1. For growth at rate r it is 1 + r, and for decay at rate r the decay factor is 1 − r. A base greater than 1 means exponential growth; a base between 0 and 1 means exponential decay. Compound interest follows A = P(1 + r ⁄ n)^(nt), and continuous compounding follows A = Pe^(rt), where e ≈ 2.71828.
  • Course examples to rehearse: 3 · 2ˣ gives 0.375, 0.75, 1.5, 3, 6, 12, 24 at x = −3 to 3. The 100-gram sample 100 · (1 ⁄ 2)ᵗ leaves 12.5 grams after three half-lives. 1,000 dollars at 6% compounded annually grows to 1,123.60 dollars in two years, against 1,120.00 dollars with simple interest. 1,500 dollars at 5% compounded continuously for 4 years gives about 1,832.10 dollars.
  • Predictable mistakes → cures: (1) evaluating 3 · 2ˣ as 6ˣ. Cure: the exponent goes first, so 3 · 2² = 12, not 36. (2) calling 3 · 2ˣ decay because f(−2) = 0.75 is small. Cure: the base decides, and 2 > 1. (3) writing b = 0.03 for 3% growth. Cure: add the rate to 1, so b = 1.03. (4) drawing the graph through the x-axis. Cure: with a > 0, every output is positive, and y = 0 is an asymptote.

Week 14 — Logarithmic functions

  • Key idea: the log is the exponent you need.
  • Definitions: log_b(x) = y means bʸ = x, with base b positive and not 1 and argument x positive. The common logarithm log(x) has base 10, and the natural logarithm ln(x) has base e. The argument must be positive, which gives every log graph a vertical asymptote where the argument is zero. The three properties are product, log_b(MN) = log_b(M) + log_b(N); quotient, log_b(M ⁄ N) = log_b(M) − log_b(N); and power, log_b(Mᵖ) = p · log_b(M).
  • Course examples to rehearse: log₂(8) = 3, log₂(1024) = 10, log₂(1⁄4) = −2, log(1000) = 3, and ln(e) = 1. log₃(x − 2) has domain (2, ∞) and asymptote x = 2. log_b(x³y²) = 3 log_b(x) + 2 log_b(y), and 2 log(x) + log(y) = log(x²y).
  • Predictable mistakes → cures: (1) splitting log(x + 4) into log(x) + log(4). Cure: there is no rule for the log of a sum; log₂(4 + 4) = 3, but log₂(4) + log₂(4) = 4. (2) log(100) = 10. Cure: ask which power of 10 gives 100; it is 2. (3) giving a real value for log₂(−8). Cure: every power of 2 is positive, so it has none. (4) drawing a log graph across its vertical asymptote. Cure: the argument must stay positive.

Week 15 — Exponential and logarithmic equations

  • Key idea: to get a variable down from an exponent, match the bases or take a logarithm of both sides. To get it out of a logarithm, convert to exponential form and check the domain.
  • Definitions: the same-base property says that if bᵐ = bⁿ, then m = n. To take a logarithm of both sides is to apply the same log, ln or log, to each side so the power property brings the exponent down. The domain check tests each candidate in every original argument and rejects any that makes an argument zero or negative. The doubling time of continuous growth is ln 2 ⁄ k and the tripling time is ln 3 ⁄ k, whatever the starting amount. The rule of 72 estimates a doubling time as 72 divided by the percent rate.
  • Course examples to rehearse: 2ˣ = 16 gives x = 4, and 9ˣ = 3^(x + 4) gives x = 4. 2ˣ = 10 gives x = ln 10 ⁄ ln 2 ≈ 3.3219. log₂(x − 1) = 3 gives x = 9. log(x) + log(x − 3) = 1 gives x = 5 only. At 6% continuous, money doubles in about 11.55 years, and the rule of 72 estimates 12. The decay A = A₀ · e^(−0.15t) has a half-life of about 4.62 hours.
  • Predictable mistakes → cures: (1) solving 2ˣ = 16 as 2x = 16. Cure: x is an exponent, and 2⁸ = 256, not 16. (2) keeping x = −2 in log(x) + log(x − 3) = 1. Cure: run the domain check; log(−2) has no real value. (3) thinking a bigger deposit doubles faster. Cure: the starting amount cancels. (4) mixing ln and log in one solve. Cure: use the same logarithm on both sides.

Representative practice

These are fresh variants, not the final's questions. Work each worked example yourself first, then compare your steps. Every number below is computed and checked in advance.

Worked example — six first-half moves (Weeks 1–7)

A campus tutoring center posts a six-line warm-up board before finals week.

  1. Simplify (3a²b)² · a⁻³. Power of a product: (3a²b)² = 9a⁴b². Then 9a⁴b² · a⁻³ = 9a⁴⁻³b² = 9ab². Check at a = 2 and b = 1: (3 · 4 · 1)² = 144, and 144 ⁄ 8 = 18; also 9 · 2 · 1 = 18.
  2. Solve 5 − 3x ≤ 20. Subtract 5: −3x ≤ 15. Divide by −3 and flip: x ≥ −5, written [−5, ∞). Check: at x = −5, 5 + 15 = 20, and 20 ≤ 20 is true.
  3. Compose with f(x) = x² + 1 and g(x) = 2x − 3. (f ∘ g)(2) = f(1) = 2, but (g ∘ f)(2) = g(5) = 7. Inside first, and the order matters.
  4. Find a line through (−2, 1) and (4, 4). m = (4 − 1) ⁄ (4 − (−2)) = 3 ⁄ 6 = 1⁄2. Point-slope form gives y − 1 = (1⁄2)(x + 2), so y = (1⁄2)x + 2. A perpendicular line has slope −2.
  5. Solve a system: 2x + y = 11 and x − y = 1. Add the equations: 3x = 12, so x = 4 and y = 3. Check: 8 + 3 = 11 and 4 − 3 = 1.
  6. Factor and solve 2x² − 5x − 3 = 0. Here ac = −6, and −6 and 1 multiply to −6 and add to −5. Split and group: 2x² − 6x + x − 3 = 2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3). So x = −1⁄2 or x = 3. The discriminant 25 + 24 = 49 is positive, which predicted two real solutions.

Why it matters: each line is one core move from Week 8's list. Naming the move first is what keeps these from blurring together on a 40-question exam.

Worked example — the best price (Week 9)

A bike-rental shop finds that at a price of p dollars a day it rents 120 − 4p bikes. Its daily revenue is R(p) = p(120 − 4p) = −4p² + 120p dollars.

  1. Opening: a = −4 is negative, so the parabola opens down, and the vertex is a maximum.
  2. Vertex: h = −b ⁄ (2a) = −120 ⁄ (2 · (−4)) = −120 ⁄ (−8) = 15. Then R(15) = 15 · (120 − 60) = 15 · 60 = 900.
  3. Check by symmetry: R(10) = 10 · 80 = 800 and R(20) = 20 · 40 = 800. Equal heights sit the same distance from p = 15.
  4. Intercepts: R(p) = 0 when p = 0 or 120 − 4p = 0, so at p = 0 and p = 30. The axis sits halfway: (0 + 30) ⁄ 2 = 15.
  5. Vertex form: R(p) = −4(p − 15)² + 900. Check at p = 0: −4 · 225 + 900 = 0.

In words: a price of 15 dollars a day brings in the most revenue, 900 dollars a day. Charging less or more brings in less.

Now read a function in vertex form. For f(x) = 2(x − 2)² − 8, the vertex is (2, −8). Since a = 2 is positive, −8 is the minimum value, and the range is [−8, ∞). Expanding gives 2(x² − 4x + 4) − 8 = 2x² − 8x + 8 − 8 = 2x² − 8x. The discriminant is (−8)² − 4(2)(0) = 64, which is positive, so there are two x-intercepts. Factoring 2x(x − 4) = 0 gives (0, 0) and (4, 0), and the axis x = 2 sits halfway between them.

Worked example — reading a graph from its formula (Week 10)

Before you open a graphing app, predict two graphs, then let the app confirm them.

A polynomial: p(x) = −(x + 2)(x − 3)².

  1. Degree and ends: the leading terms multiply to −x · x² = −x³. The degree 3 is odd and the leading coefficient −1 is negative, so the graph rises on the left and falls on the right.
  2. Zeros: x = −2 with multiplicity 1, so the graph crosses there. x = 3 with multiplicity 2, so it touches and turns back.
  3. Check the signs: p(−3) = −(−1)(36) = 36 and p(−1) = −(1)(16) = −16, so the sign changes at −2. p(2) = −(4)(1) = −4 and p(4) = −(6)(1) = −6, so the sign stays negative around 3.
  4. The y-intercept: p(0) = −(2)(9) = −18, so the point is (0, −18).

A rational function: r(x) = (x² − 4) ⁄ (x² − x − 6).

  1. Factor first: r(x) = (x − 2)(x + 2) ⁄ ((x − 3)(x + 2)). The domain leaves out x = 3 and x = −2.
  2. Hole or asymptote: the factor x + 2 cancels, so x = −2 is a hole. The simplified form (x − 2) ⁄ (x − 3) gives its height: (−4) ⁄ (−5) = 4⁄5, so the hole is at (−2, 4⁄5). The factor x − 3 remains, so x = 3 is a vertical asymptote.
  3. Horizontal asymptote: the degrees are equal, 2 and 2, so y = 1 ⁄ 1 = 1.
  4. Intercepts: the x-intercept is (2, 0), and r(0) = −4 ⁄ (−6) = 2⁄3 gives (0, 2⁄3).
  5. Far out: at x = 103 the simplified form is 101 ⁄ 100 = 1.01, close to the asymptote y = 1.

Why it matters: calling x = −2 an asymptote because it makes the denominator 0 skips the factoring step that tells a hole from an asymptote.

Worked example — fractions with polynomials (Week 11)

Simplify (x² − 16) ⁄ (x² + x − 12).

  1. Excluded values first: x² + x − 12 = (x + 4)(x − 3), which is 0 at x = −4 and x = 3.
  2. Factor and cancel: (x − 4)(x + 4) ⁄ ((x + 4)(x − 3)) = (x − 4) ⁄ (x − 3), for x ≠ −4 and x ≠ 3.
  3. Check at x = 0: the original is −16 ⁄ (−12) = 4⁄3, and the simplified form is (−4) ⁄ (−3) = 4⁄3.

Add 3 ⁄ (x² − 1) + 1 ⁄ (x + 1). Factor the first denominator: x² − 1 = (x − 1)(x + 1), so that product is the LCD. Rewrite the second fraction as (x − 1) ⁄ ((x − 1)(x + 1)). The sum is (3 + x − 1) ⁄ ((x − 1)(x + 1)) = (x + 2) ⁄ ((x − 1)(x + 1)), for x ≠ 1 and x ≠ −1. Check at x = 0: 3 ⁄ (−1) + 1 ⁄ 1 = −2, and 2 ⁄ (−1) = −2.

Solve a sharing problem. Coworkers split the cost of a 480-dollar group gift equally. If 4 more people join, each person pays 20 dollars less. With n people at first, 480 ⁄ n − 480 ⁄ (n + 4) = 20, where n ≠ 0 and n ≠ −4.

  1. Multiply every term by n(n + 4): 480(n + 4) − 480n = 20n(n + 4).
  2. Simplify: 480n + 1920 − 480n = 20n² + 80n, so 1920 = 20n² + 80n.
  3. Divide by 20: 96 = n² + 4n, so n² + 4n − 96 = 0.
  4. Factor: (n + 12)(n − 8) = 0, so n = −12 or n = 8. A number of people cannot be negative, so n = 8.
  5. Check: 480 ⁄ 8 = 60 dollars each and 480 ⁄ 12 = 40 dollars each, and 60 − 40 = 20.

Worked example — a tile, a power, and a check (Week 12)

A mosaic class cuts a square tile with an area of 98 square inches. Its side is √98 inches.

  1. Simplify: the largest perfect-square factor of 98 is 49, so √98 = √(49 · 2) = 7√2.
  2. Check: (7√2)² = 49 · 2 = 98. As a decimal, 7√2 ≈ 7 × 1.4142 ≈ 9.899 inches.
  3. The sum trap: 98 = 49 + 49, but √(49 + 49) is not 7 + 7 = 14. The side is about 9.899 inches.

A rational exponent: 49^(3⁄2) = (√49)³ = 7³ = 343, because 7² = 49. Treating the exponent as a multiplier gives 49 × 3⁄2 = 73.5, which is wrong.

The exponent rules with fractions: x^(2⁄5) · x^(3⁄10) = x^(4⁄10 + 3⁄10) = x^(7⁄10). Check at x = 1024, which is 2¹⁰: 1024^(2⁄5) · 1024^(3⁄10) = 2⁴ · 2³ = 16 · 8 = 128, and 1024^(7⁄10) = 2⁷ = 128.

A radical equation: √(2x + 1) + 1 = x.

  1. Isolate the radical: √(2x + 1) = x − 1.
  2. Square both sides: 2x + 1 = (x − 1)² = x² − 2x + 1. Keep the middle term −2x.
  3. Solve: 0 = x² − 4x = x(x − 4), so x = 0 or x = 4.
  4. Check x = 4: √9 + 1 = 3 + 1 = 4. It works.
  5. Check x = 0: √1 + 1 = 2, and 2 ≠ 0. Reject x = 0 as an extraneous solution. The only solution is x = 4.

Worked example — growth, decay, and interest (Week 13)

A neighborhood seed library starts with 250 packets and grows 20% per year. The growth factor is 1 + 0.20 = 1.2, so P(t) = 250 · (1.2)ᵗ. After one year P(1) = 250 · 1.2 = 300 packets, and after two years P(2) = 300 · 1.2 = 360 packets. Writing b = 0.2 instead would shrink the library to a fifth of its size each year.

Now a decay function: f(x) = 6 · (1⁄3)ˣ. Its base 1⁄3 is between 0 and 1, so it decays. f(0) = 6 is the initial value, f(1) = 2, and f(2) = 6 ⁄ 9 = 2⁄3. A negative input gives a larger output, f(−1) = 6 · 3 = 18, but never a negative one. The graph crosses the y-axis at (0, 6) and flattens toward y = 0 on the right.

Compound interest. A credit union pays 8% compounded quarterly on 5,000 dollars for one year. Each quarter pays r ⁄ n = 0.08 ⁄ 4 = 0.02, and there are nt = 4 · 1 = 4 quarters.

  1. After quarter 1: 5000 · 1.02 = 5,100.00 dollars.
  2. After quarter 2: 5100 · 1.02 = 5,202.00 dollars.
  3. After quarter 3: 5202 · 1.02 = 5,306.04 dollars.
  4. After quarter 4: 5306.04 · 1.02 = 5,412.1608, or 5,412.16 dollars to the cent.

The formula agrees: 5000 · (1.02)⁴ = 5000 · 1.08243216 ≈ 5,412.16 dollars. Compounded once a year it would be 5000 · 1.08 = 5,400.00 dollars, so quarterly compounding earns 12.16 dollars more.

Worked example — logs as exponents (Week 14)

  1. Evaluate: log₆(36) = 2 because 6² = 36. log₄(2) = 1⁄2 because 4^(1⁄2) = √4 = 2. log(0.001) = −3 because 10⁻³ = 1 ⁄ 1000. On a base-10 scale, a ratio of 1 ⁄ 1000 sits three steps down.
  2. Domain: log₂(5 − x) needs 5 − x > 0, so x < 5. The domain is (−∞, 5), and the vertical asymptote is x = 5. Check: at x = 4 the value is log₂(1) = 0, and at x = −3 it is log₂(8) = 3.
  3. Expand: log₂(4x² ⁄ y) = log₂(4) + 2 log₂(x) − log₂(y) = 2 + 2 log₂(x) − log₂(y). Check at x = 2 and y = 4: 4 · 4 ⁄ 4 = 4 and log₂(4) = 2, while 2 + 2 · 1 − 2 = 2.
  4. Condense: 2 log(x) + log(4) − log(y) = log(4x² ⁄ y). Check at x = 10 and y = 4: the left side is 2 + log(4) − log(4) = 2, and log(4 · 100 ⁄ 4) = log(100) = 2.
  5. The sum trap: log₂(8 + 8) = log₂(16) = 4, but log₂(8) + log₂(8) = 3 + 3 = 6.

Worked example — solving for the exponent (Week 15)

  1. Match the bases: 25ˣ = 125^(x − 1). Rewrite with base 5: 5^(2x) = 5^(3x − 3). Match the exponents: 2x = 3x − 3, so x = 3. Check: 25³ = 15,625 and 125² = 15,625.
  2. Take a log of both sides: 3ˣ = 10. Since 3² = 9 and 3³ = 27, x is between 2 and 3. Take ln: x · ln 3 = ln 10, so x = ln 10 ⁄ ln 3. With the given constants, x ≈ 2.3026 ⁄ 1.0986 ≈ 2.096. The exact form is the result; the decimal is a check.
  3. Condense, convert, check: log₂(x) + log₂(x + 2) = 3. The domain needs x > 0. Condense: log₂(x(x + 2)) = 3, so x² + 2x = 8. Then x² + 2x − 8 = 0 and (x + 4)(x − 2) = 0, so x = −4 or x = 2. Only x = 2 passes the domain check. Check: log₂(2) + log₂(4) = 1 + 2 = 3.
  4. Doubling time: a scholarship fund earns 9% compounded continuously. It doubles when e^(0.09t) = 2, so t = ln 2 ⁄ 0.09 ≈ 0.6931 ⁄ 0.09 ≈ 7.70 years. The rule of 72 estimates 72 ⁄ 9 = 8 years. The fund's size never enters the calculation.
  5. Half-life: a lab tracer follows A = A₀ · e^(−0.25t), with t in hours, so k = −0.25. The half-life is ln 2 ⁄ 0.25 ≈ 0.6931 ⁄ 0.25 ≈ 2.77 hours, and two half-lives take about 5.54 hours.

Self-check questions

Try each one, then read the check and the one-line why.

  1. Simplify (2x⁻³)⁻². Check: x⁶ ⁄ 4. Why: the power rule gives 2⁻² · x⁶, and 2⁻² = 1⁄4.
  2. Write the solutions of |3x + 1| < 8 in interval notation. Check: (−3, 7⁄3). Why: less than means the bounded middle, −8 < 3x + 1 < 8, so −9 < 3x < 7.
  3. Find the domain of √(6 − 2x). Check: (−∞, 3]. Why: the radicand must be zero or more, and 6 − 2x ≥ 0 gives x ≤ 3.
  4. Factor 4x² − 49. Check: (2x + 7)(2x − 7). Why: it is a difference of squares, (2x)² − 7².
  5. Find the vertex of f(x) = −3x² − 12x + 1. Is it a maximum or a minimum? Check: (−2, 13), a maximum. Why: h = 12 ⁄ (−6) = −2, k = −12 + 24 + 1 = 13, and a = −3 is negative.
  6. Describe the end behavior of 5x − 2x³ + 1. Check: rises on the left, falls on the right. Why: the leading term −2x³ has an odd degree and a negative coefficient.
  7. Find the asymptotes of (6x² − 1) ⁄ (2x² + 5x). Check: horizontal y = 3; vertical x = 0 and x = −5⁄2. Why: equal degrees give 6 ⁄ 2 = 3, and 2x² + 5x = x(2x + 5) with no factor cancelling.
  8. Is (x² − 25) ⁄ (x − 5) the same function as x + 5? Check: no. Why: the quotient is x + 5 only for x ≠ 5, so its graph has a hole at (5, 10).
  9. Compute 9^(3⁄2) and 4^(−1⁄2). Check: 27 and 1⁄2. Why: root first, (√9)³ = 3³ = 27, and a negative exponent gives the reciprocal of √4 = 2.
  10. Growth or decay: g(x) = 0.5 · 3ˣ and h(x) = 7 · (0.9)ˣ? Check: g grows and h decays. Why: the base decides, 3 > 1 and 0.9 is between 0 and 1; the initial values 0.5 and 7 do not.
  11. Evaluate log₅(1) and log₉(3). Check: 0 and 1⁄2. Why: 5⁰ = 1, and 9^(1⁄2) = √9 = 3.
  12. Solve log₅(2x + 1) = 2. Check: x = 12. Why: 5² = 2x + 1 gives 2x = 24, and log₅(25) = 2 checks.
  13. How long does money take to triple at 10% compounded continuously? Check: about 10.99 years. Why: ln 3 ⁄ 0.1 ≈ 1.0986 ⁄ 0.1 ≈ 10.99.

Check yourself — choose one. Write two or three sentences to a friend.

  • (a) Explain what the bike shop's vertex (15, 900) means in the story, and why charging 20 dollars brings in less.
  • (b) Explain why x = 0 in the radical equation is rejected even though every algebra step that produced it was correct.

Either one rehearses an explain-why skill the final checks.

Watch before the exam

These seven videos come from the course's own weekly readings, and all seven have captions. Use them to review and to connect weeks, not to learn anything new.

Study plan

A short countdown, counted back from the day you plan to sit the final in Week 16. Start it about a week ahead, while Week 15's material is still fresh.

  • As soon as the practice exam opens: take the practice exam once, cold and closed-note. Sort every miss by week. Remember that it leans on Weeks 1–7 far more than the final does.
  • About six days before you sit the final: review Weeks 10 and 11 above and redo those two worked examples from scratch, without looking.
  • About five days before: review Week 12 and redo the tile, the power, and the radical equation.
  • About four days before: review Weeks 13 and 14 and redo their worked examples. Say each log value as an exponent sentence: "6 squared is 36, so log base 6 of 36 is 2."
  • About three days before: review Week 15 and the Week 9 parabola. Then run the paired exam-prep tutorial (the copy-and-paste chatbot tutor, or its no-AI route) on the two areas that felt weakest, with at least one from Weeks 10–15.
  • About two days before: skim the Weeks 1–6 toolkit, redo the six-move warm-up board, and work the thirteen self-check questions cold.
  • The day before: retake the practice exam, redo every miss by hand, and watch one video from the list above. Then skim every "predictable mistakes → cures" line and stop studying.

This guide is for review. The exam-prep tutorial and the practice exam are where you rehearse. Retrieval practice, not rereading, is what moves scores.

Exam format

The final is scored automatically, and all 40 questions count the same. The final checks your own work without AI. The questions come one at a time, and there is no going back, so settle each one before you move on.

Test-taking strategies for this material:

  • Name the move before you start. Is it simplify, solve, read a graph's features, simplify a fraction, take a root, model growth, evaluate a log, or solve for an exponent?
  • Factor first in every rational question. Then sort each excluded value into a hole or an asymptote.
  • List the excluded values and the domain before you solve a rational, radical, or logarithmic equation. Test every candidate against them at the end.
  • Read the leading term alone for end behavior, and the exponent on each factor for crossing or touching.
  • Let the base decide growth or decay, and add the rate to 1 for the growth factor.
  • Say every log as an exponent sentence, and refuse to split the log of a sum.
  • Keep exact forms. Leave ln 10 ⁄ ln 3 or 7√2 as written, and use a decimal only as a check.
  • Use scratch paper for every computation, even a short one. Writing each step catches most dropped signs.

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