Sequences, Exponential, and Logarithmic Functions Comprehensive Study Notes

Course Schedule and Curriculum Overview: Unit 2 (Fall 2026)

  • Course Topic: Unit 2 — Sequences, Exponential and Logarithmic Functions

  • Student / Author Record: Keegan Whitaker, Dated 8/31/26

  • Fall 2026 Daily Schedule and Topic Breakdown:

    • Monday, 8/31: Exponential Expressions and Manipulations (Exponent Rules) — Worksheet #1

    • Tuesday, 9/1: Exponential Expressions and Manipulations Continued

    • Wednesday, 9/2: Arithmetic and Geometric Sequences with Focus on Rate of Change (ROC), Convergence, and Divergence — Worksheet #2

    • Thursday, 9/3: Exponential Functions with Transformations, End Behavior, and Exponential in Context (Best Fit) — Worksheet #3

    • Friday, 9/4: Residuals — Worksheet #4

    • Tuesday, 9/8: Review and Practice — Worksheet #5

    • Wednesday, 9/9: Quiz #1

    • Thursday, 9/10: Inverses and Compositions — Worksheet #6

    • Friday, 9/11: Review and Free Response Question (FRQ) #2

    • Monday, 9/14: Inverses of Exponential Functions — Worksheet #7

    • Tuesday, 9/15: Quiz #2

    • Wednesday, 9/16: Introduction to Logarithmic Functions and Graphing Logarithmic Functions with Characteristics (including End Behavior) — Worksheet #8

    • Thursday, 9/17: DeltaMath Quiz Review Opens

    • Friday, 9/18: Properties of Logarithms and Review — Worksheet #9

    • Monday, 9/28: Solving Logarithmic and Exponential Equations — Worksheet #10

    • Tuesday, 9/29: Solving Logarithmic and Exponential Inequalities — Worksheet #11

    • Wednesday, 9/30: Applications and Graphing with Semi-Log Paper — Worksheet #12

    • Fall Break Block: Unit 2 Test, AP Classroom Review, Optional DeltaMath Test Review/Classwork, Notebook Submissions Due

Exponent Rules, Rational Exponents, and Expression Manipulations

  • Core Mathematical Rules for Exponents and Radicals:

    • Rational Exponent Definition: a^{m/n} = \n\sqrt[n]{a^m} = (\sqrt[n]{a})^m

    • Negative Exponent Rule: a−n=1ana^{-n} = \frac{1}{a^n}

    • Product of Powers Rule: am⋅an=am+na^m \cdot a^n = a^{m+n}

    • Quotient of Powers Rule: aman=am−n\frac{a^m}{a^n} = a^{m-n}

    • Power of a Power Rule: (am)n=am⋅n(a^m)^n = a^{m \cdot n}

    • Power of a Product Rule: (ab)n=anbn(ab)^n = a^n b^n

    • Power of a Quotient Rule: (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

  • Numerical Simplifications Without Calculators:

    • (81)3/4=(814)3=33=27(81)^{3/4} = (\sqrt[4]{81})^3 = 3^3 = 27

    • (16)3/4=(164)3=23=8(16)^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8

    • (32)2/5=(325)2=22=4(32)^{2/5} = (\sqrt[5]{32})^2 = 2^2 = 4

    • (27)2/3=(273)2=32=9(27)^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9

    • (512)−2/3=1(5123)2=182=164(512)^{-2/3} = \frac{1}{(\sqrt[3]{512})^2} = \frac{1}{8^2} = \frac{1}{64}

    • (4125)−2/3=(1254)2/3\left(\frac{4}{125}\right)^{-2/3} = \left(\frac{125}{4}\right)^{2/3}; or (12564)−2/3=(64125)2/3=1625\left(\frac{125}{64}\right)^{-2/3} = \left(\frac{64}{125}\right)^{2/3} = \frac{16}{25}

  • Variable and Radical Expression Manipulations:

    • 5a3/15a3/2=5a1/5a3/2=5a1/5−3/2=5a−13/10=5a13/10\frac{5a^{3/15}}{a^{3/2}} = \frac{5a^{1/5}}{a^{3/2}} = 5a^{1/5 - 3/2} = 5a^{-13/10} = \frac{5}{a^{13/10}}

    • 12x1/2x2/3=12x1/2−2/3=12x−1/6=12x1/6\frac{12x^{1/2}}{x^{2/3}} = 12x^{1/2 - 2/3} = 12x^{-1/6} = \frac{12}{x^{1/6}}

    • 2x2⋅3x−2/36x1/6=6x4/36x1/6=x4/3−1/6=x7/6\frac{2x^2 \cdot 3x^{-2/3}}{6x^{1/6}} = \frac{6x^{4/3}}{6x^{1/6}} = x^{4/3 - 1/6} = x^{7/6}

    • 4e3y1/2e5x=4e3−5xy1/2\frac{4e^3 y^{1/2}}{e^{5x}} = 4e^{3-5x} y^{1/2}

    • (−4e2x)3=(−4)3e6x=−64e6x(-4e^{2x})^3 = (-4)^3 e^{6x} = -64e^{6x}

    • 6x⋅e6x3e4x=2xe2x\frac{6x \cdot e^{6x}}{3e^{4x}} = 2x e^{2x}

    • e5x−24e3=14e5x−5\frac{e^{5x-2}}{4e^3} = \frac{1}{4}e^{5x-5}

    • 16e10x=4e5x\sqrt{16e^{10x}} = 4e^{5x}

    • (2e−2x)−1=12e2x(2e^{-2x})^{-1} = \frac{1}{2} e^{2x}

    • 16e12x=16e6x16\sqrt{e^{12x}} = 16e^{6x}

    • xy2(xy2)−1=1xy^2(xy^2)^{-1} = 1

    • a2b3(ab)2(a2b3)1/2=a2b3a2b2a1b3/2=a5b13/2a^2 b^3 (ab)^2 (a^2 b^3)^{1/2} = a^2 b^3 a^2 b^2 a^1 b^{3/2} = a^5 b^{13/2}

    • 2(64x6)2/3=2(4x2)2=32x42(64x^6)^{2/3} = 2(4x^2)^2 = 32x^4

    • 35−x(27x2)3=35−x(33x2)3=35−x⋅39x6=314−xx63^{5-x} (27x^2)^3 = 3^{5-x} (3^3 x^2)^3 = 3^{5-x} \cdot 3^9 x^6 = 3^{14-x} x^6

    • [(2x+3)(2x−3)]2=(4x2−9)2=16x4−72x2+81[(2x+3)(2x-3)]^2 = (4x^2 - 9)^2 = 16x^4 - 72x^2 + 81

    • x(x+1)(x−1)=x(x2−1)=x3−xx(x+1)(x-1) = x(x^2 - 1) = x^3 - x

    • (x+y)2x+y=x+y\frac{(x+y)^2}{x+y} = x+y

    • 9n+19n−1=9(n+1)−(n−1)=92=81\frac{9^{n+1}}{9^{n-1}} = 9^{(n+1)-(n-1)} = 9^2 = 81

    • a2n−1an−2=a(2n−1)−(n−2)=an+1\frac{a^{2n-1}}{a^{n-2}} = a^{(2n-1)-(n-2)} = a^{n+1}

    • (49n3m77m1n3)−2=(7m6)−2=149m12\left(\frac{49n^3 m^7}{7m^1 n^3}\right)^{-2} = (7m^6)^{-2} = \frac{1}{49m^{12}}

    • 3x+1+3x+13x+1+3x+1+3x+1=2⋅3x+13⋅3x+1=23\frac{3^{x+1}+3^{x+1}}{3^{x+1}+3^{x+1}+3^{x+1}} = \frac{2 \cdot 3^{x+1}}{3 \cdot 3^{x+1}} = \frac{2}{3}

    • (16)−2x(136)−3=62x⋅(6−2)−3=62x⋅66=62x+6\left(\frac{1}{6}\right)^{-2x} \left(\frac{1}{36}\right)^{-3} = 6^{2x} \cdot (6^{-2})^{-3} = 6^{2x} \cdot 6^6 = 6^{2x+6}

  • Simplifications Rewritten with Single Positive Exponent:

    • (2−3)4=2−12=1212(2^{-3})^4 = 2^{-12} = \frac{1}{2^{12}}

    • (a−2)3=a−6=1a6(a^{-2})^3 = a^{-6} = \frac{1}{a^6}

    • (b−3)0=b0=1(b^{-3})^0 = b^0 = 1

    • (m−1)−2=m2(m^{-1})^{-2} = m^2

  • Simplifications Involving Radicals and Absolute Values:

    • −(c−7d)2=−∣c−7d∣-\sqrt{(c-7d)^2} = -|c-7d|

    • ±49(e−f)2=±7∣e−f∣\pm \sqrt{49(e-f)^2} = \pm 7|e-f|

    • (g+3)2(g−3)2=∣(g+3)(g−3)∣=∣g2−9∣\sqrt{(g+3)^2(g-3)^2} = |(g+3)(g-3)| = |g^2 - 9|

    • 4a2−20a+25=(2a−5)2=∣2a−5∣\sqrt{4a^2 - 20a + 25} = \sqrt{(2a-5)^2} = |2a-5|

    • 1−6c+9c2=(1−3c)2=∣1−3c∣\sqrt{1-6c+9c^2} = \sqrt{(1-3c)^2} = |1-3c|

  • Rational Algebraic Simplifications:

    • m2−2m−154m−20=(m−5)(m+3)4(m−5)=m+34\frac{m^2-2m-15}{4m-20} = \frac{(m-5)(m+3)}{4(m-5)} = \frac{m+3}{4}

Exponential Functions, Transformations, and End Behavior

  • Parent Function Characteristics and Key Points:

    • y=2xy = 2^x: Key points at (−2,0.25)(-2, 0.25), (−1,0.5)(-1, 0.5), (0,1)(0, 1), (1,2)(1, 2), (2,4)(2, 4); Horizontal Asymptote at y=0y = 0.

    • y=3xy = 3^x: Key points at (−2,1/9)(-2, 1/9), (−1,1/3)(-1, 1/3), (0,1)(0, 1), (1,3)(1, 3), (2,9)(2, 9); Horizontal Asymptote at y=0y = 0.

    • y=4x−1y = 4^{x-1}: Horizontal shift right by 11 unit. Key points at (0,1/4)(0, 1/4), (1,1)(1, 1), (2,4)(2, 4), (3,16)(3, 16); Horizontal Asymptote at y=0y = 0.

    • y=5x+2y = 5^{x+2}: Horizontal shift left by 22 units. Key points at (−2,1)(-2, 1), (−1,5)(-1, 5), (0,25)(0, 25); Horizontal Asymptote at y=0y = 0.

    • y=(14)xy = \left(\frac{1}{4}\right)^x: Exponential decay graph. Key points at (−2,16)(-2, 16), (−1,4)(-1, 4), (0,1)(0, 1), (1,1/4)(1, 1/4); Horizontal Asymptote at y=0y = 0.

    • y=−4xy = -4^x: Vertical reflection across the x-axis. Key points at (0,−1)(0, -1), (1,−4)(1, -4), (2,−16)(2, -16); Horizontal Asymptote at y=0y = 0.

    • y=−(13)xy = -\left(\frac{1}{3}\right)^x: Vertical reflection across the x-axis. Key points at (−2,−9)(-2, -9), (−1,−3)(-1, -3), (0,−1)(0, -1), (1,−1/3)(1, -1/3); Horizontal Asymptote at y=0y = 0.

    • y=−(15)xy = -\left(\frac{1}{5}\right)^x: Vertical reflection across the x-axis. Key points at (−1,−5)(-1, -5), (0,−1)(0, -1), (1,−1/5)(1, -1/5); Horizontal Asymptote at y=0y = 0.

  • Evaluated Function Tables:

    • For y=3(5)xy = 3(5)^x:

      • When x=−2x = -2, y=3/25=0.12y = 3/25 = 0.12

      • When x=0x = 0, y=3y = 3

      • When x=3x = 3, y=375y = 375

    • For y=−2(9)xy = -2(9)^x:

      • When x=0x = 0, y=−2y = -2

      • When x=1/2x = 1/2, y=−2(3)=−6y = -2(3) = -6

      • When x=1x = 1, y=−18y = -18

      • When x=2x = 2, y=−162y = -162

    • For y=(0.5)xy = (0.5)^x:

      • When x=−2x = -2, y=4y = 4

      • When x=−1x = -1, y=2y = 2

      • When x=0x = 0, y=1y = 1

      • When x=1x = 1, y=0.5y = 0.5

      • When x=2x = 2, y=0.25y = 0.25

  • Transformation Descriptions from Parent Functions:

    • y=2x−4y = 2^{x-4}: Shifted right 44 units.

    • y=12(4x)y = \frac{1}{2}(4^x): Vertically compressed by a factor of 12\frac{1}{2}.

    • y=3(2x+1)−8y = 3(2^{x+1}) - 8: Vertically stretched by a factor of 33, shifted left 11 unit, shifted down 88 units. Horizontal Asymptote at y=−8y = -8.

    • y=−5(3x+4)+5y = -5(3^{x+4}) + 5: Reflected across the x-axis, vertically stretched by a factor of 55, shifted left 44 units, shifted up 55 units. Horizontal Asymptote at y=5y = 5.

    • y=3x−2−10y = 3^{x-2} - 10: Shifted right 22 units, shifted down 1010 units. Horizontal Asymptote at y=−10y = -10.

    • y=22x−8y = 2^{2x} - 8: Horizontally compressed by a factor of 12\frac{1}{2}, shifted down 88 units. Horizontal Asymptote at y=−8y = -8.

  • Growth/Decay Classification and Limit End Behavior:

    • f(x)=3−2x=(19)xf(x) = 3^{-2x} = \left(\frac{1}{9}\right)^x: Exponential Decay.

      • lim⁡x→∞f(x)=0\lim_{x \to \infty} f(x) = 0

      • lim⁡x→−∞f(x)=∞\lim_{x \to -\infty} f(x) = \infty

    • g(x)=2(0.55)xg(x) = 2(0.55)^x: Exponential Decay.

      • lim⁡x→∞g(x)=0\lim_{x \to \infty} g(x) = 0

      • lim⁡x→−∞g(x)=∞\lim_{x \to -\infty} g(x) = \infty

    • k(x)=3x−1k(x) = 3^{x-1}: Exponential Growth.

      • lim⁡x→∞k(x)=∞\lim_{x \to \infty} k(x) = \infty

      • lim⁡x→−∞k(x)=0\lim_{x \to -\infty} k(x) = 0

  • Demonstration of Function Equivalence:

    • Showing y1=32x+4y_1 = 3^{2x+4} is identical to y2=9x+2y_2 = 9^{x+2}.

      • y2=9x+2=(32)x+2=32(x+2)=32x+4=y1y_2 = 9^{x+2} = (3^2)^{x+2} = 3^{2(x+2)} = 3^{2x+4} = y_1

    • Showing y1=2(23x−2)y_1 = 2(2^{3x-2}) is identical to y2=23x−1y_2 = 2^{3x-1}.

      • y1=21⋅23x−2=21+(3x−2)=23x−1=y2y_1 = 2^1 \cdot 2^{3x-2} = 2^{1 + (3x-2)} = 2^{3x-1} = y_2

    • Showing e3+e51+e2\frac{e^3 + e^5}{1 + e^2} is identical to e3e^3.

      • e3(1+e2)1+e2=e3\frac{e^3(1+e^2)}{1+e^2} = e^3

  • Natural Exponential Functions:

    • Parent natural exponential function: y=exy = e^x

    • Transformed natural exponential function: y=2ex−3+1y = 2e^{x-3} + 1

      • Transformations: Vertically stretched by 22, shifted right 33 units, shifted up 11 unit. Horizontal asymptote at y=1y = 1

Exponential Growth, Decay, and Real-World Modeling

  • Identifying Constant Percentage Growth and Decay Rates:

    • P(t)=3.5(1.09)tP(t) = 3.5(1.09)^t: Exponential Growth; constant rate r=9%r = 9\% (0.090.09

    • P(t)=4.3(1.018)tP(t) = 4.3(1.018)^t: Exponential Growth; constant rate r=1.8%r = 1.8\% (0.0180.018

    • f(x)=78.963(0.968)xf(x) = 78.963(0.968)^x: Exponential Decay; constant rate r=3.2%r = 3.2\% (1−0.968=0.0321 - 0.968 = 0.032

    • f(x)=5607(0.9968)xf(x) = 5607(0.9968)^x: Exponential Decay; constant rate r=0.32%r = 0.32\% (1−0.9968=0.00321 - 0.9968 = 0.0032

    • g(t)=247(2)tg(t) = 247(2)^t: Exponential Growth; constant rate r=100%r = 100\% (2−1=1.002 - 1 = 1.00

    • g(t)=43(0.05)tg(t) = 43(0.05)^t: Exponential Decay; constant rate r=95%r = 95\% (1−0.05=0.951 - 0.05 = 0.95

  • Constructing Exponential Functions y=a(1±r)ty = a(1 \pm r)^t from Conditions:

    • Initial value =5= 5, increasing at 17%17\% per year: y=5(1+0.17)t=5(1.17)ty = 5(1 + 0.17)^t = 5(1.17)^t

    • Initial value =16= 16, decreasing at 50%50\% per month: y=16(1−0.50)t=16(0.5)ty = 16(1 - 0.50)^t = 16(0.5)^t

    • Initial value =5= 5, decreasing at 0.59%0.59\% per week: y=5(1−0.0059)t=5(0.9941)ty = 5(1 - 0.0059)^t = 5(0.9941)^t

    • Initial height =18 cm= 18\,cm, growing at 5.2%5.2\% per week: y=18(1+0.052)t=18(1.052)ty = 18(1 + 0.052)^t = 18(1.052)^t

  • Constructing Exponential Formulas from Numerical Tables:

    • Given table for f(x)f(x):

      • Data points: (−2,1.472)(-2, 1.472), (−1,1.84)(-1, 1.84), (0,2.3)(0, 2.3), (1,2.875)(1, 2.875), (2,3.59375)(2, 3.59375)

      • Constant multiplier b=2.31.84=1.25b = \frac{2.3}{1.84} = 1.25; Initial value a=2.3a = 2.3

      • Formula: f(x)=2.3(1.25)xf(x) = 2.3(1.25)^x

    • Given table for g(x)g(x):

      • Data points: (−2,−9.0625)(-2, -9.0625), (−1,−7.25)(-1, -7.25), (0,−5.8)(0, -5.8), (1,−4.64)(1, -4.64), (2,−3.7123)(2, -3.7123)

      • Constant multiplier b=−4.64−5.8=0.8b = \frac{-4.64}{-5.8} = 0.8; Initial value a=−5.8a = -5.8

      • Formula: g(x)=−5.8(0.8)xg(x) = -5.8(0.8)^x

  • Constructing Exponential Formulas from Coordinate Graphs:

    • Graph passing through (0,4)(0,4) and (5,8.05)(5, 8.05):

      • Initial value a=4a = 4

      • Solve for bb: 8.05=4(b)5  ⟹  b5=2.0125  ⟹  b=(2.0125)1/5≈1.1508.05 = 4(b)^5 \implies b^5 = 2.0125 \implies b = (2.0125)^{1/5} \approx 1.150

      • Formula: y=4(1.150)xy = 4(1.150)^x

    • Graph passing through (0,3)(0,3) and (4,1.49)(4, 1.49):

      • Initial value a=3a = 3

      • Solve for bb: 1.49=3(b)4  ⟹  b4=0.4967  ⟹  b=(0.4967)1/4≈0.8391.49 = 3(b)^4 \implies b^4 = 0.4967 \implies b = (0.4967)^{1/4} \approx 0.839

      • Formula: y=3(0.839)xy = 3(0.839)^x

  • Real-World Modeling Applications:

    • Jacksonville, Florida Population Model:

      • In 2020 (t=0t = 0), population =736000= 736000, growing at 1.49%1.49\% per year.

      • Population equation: f(t)=736000(1.0149)tf(t) = 736000(1.0149)^t

      • Predicted population in 2050 (t=30t = 30): f(30)=736000(1.0149)30≈1147382f(30) = 736000(1.0149)^{30} \approx 1147382

      • Time to reach 10000001000000 residents: 1000000=736000(1.0149)t  ⟹  1.3587=(1.0149)t  ⟹  t=ln⁡(1.3587)ln⁡(1.0149)≈20.7≈21 yrs1000000 = 736000(1.0149)^t \implies 1.3587 = (1.0149)^t \implies t = \frac{\ln(1.3587)}{\ln(1.0149)} \approx 20.7 \approx 21\,yrs (Year 2041).

    • Radioactive Decay Model:

      • Half-life =14 days= 14\,days; Initial amount =6.6 g= 6.6\,g

      • Remaining mass equation: y=6.6(12)t/14y = 6.6\left(\frac{1}{2}\right)^{t/14}

      • Time when less than 1 g1\,g remains: 1=6.6(12)t/14  ⟹  16.6=(0.5)t/14  ⟹  t=14⋅ln⁡(1/6.6)ln⁡(0.5)≈38.1 days1 = 6.6\left(\frac{1}{2}\right)^{t/14} \implies \frac{1}{6.6} = (0.5)^{t/14} \implies t = 14 \cdot \frac{\ln(1/6.6)}{\ln(0.5)} \approx 38.1\,days

    • Bacterial Culture Growth Model:

      • Bacteria count formula: B=100e0.693tB = 100e^{0.693t}

      • Initial amount (t=0t = 0): 100 bacteria100\,bacteria

      • Time to reach 200 bacteria200\,bacteria: 200=100e0.693t  ⟹  2=e0.693t  ⟹  t=ln⁡(2)0.693≈1.00 hour200 = 100e^{0.693t} \implies 2 = e^{0.693t} \implies t = \frac{\ln(2)}{0.693} \approx 1.00\,hour

    • Carbon-14 Decay Model:

      • Mass equation: C=20e−0.0001216tC = 20e^{-0.0001216t}

      • Initial amount: 20 g20\,g

      • Half-life determination: 10=20e−0.0001216t  ⟹  0.5=e−0.0001216t  ⟹  t=ln⁡(0.5)−0.0001216≈5700 years10 = 20e^{-0.0001216t} \implies 0.5 = e^{-0.0001216t} \implies t = \frac{\ln(0.5)}{-0.0001216} \approx 5700\,years

Arithmetic and Geometric Sequences and Series

  • Arithmetic Sequences:

    • Characteristics: Constant difference / constant rate of change d=an−an−1d = a_n - a_{n-1}.

    • Explicit Formula: an=a1+(n−1)da_n = a_1 + (n-1)d or an=dn+a0a_n = dn + a_0

    • Example 1: 6,10,14,18,…6, 10, 14, 18, \dots

      • Common difference d=4d = 4

      • Tenth term a10=6+4(9)=42a_{10} = 6 + 4(9) = 42

      • Explicit rule: an=4n+2a_n = 4n + 2

    • Example 2: −4,1,6,11,…-4, 1, 6, 11, \dots

      • Common difference d=5d = 5

      • Tenth term a10=−4+5(9)=41a_{10} = -4 + 5(9) = 41

      • Explicit rule: an=5n−9a_n = 5n - 9

    • Example 3: −5,−2,1,4,…-5, -2, 1, 4, \dots

      • Common difference d=3d = 3

      • Tenth term a10=−5+3(9)=22a_{10} = -5 + 3(9) = 22

      • Explicit rule: an=3n−8a_n = 3n - 8

    • Example 4: 36,25,14,3,…36, 25, 14, 3, \dots

      • Common difference d=−11d = -11

      • Tenth term a10=36−11(9)=−63a_{10} = 36 - 11(9) = -63

      • Explicit rule: an=−11n+47a_n = -11n + 47

    • Finding terms from given elements:

      • Arithmetic sequence with a3=−8a_3 = -8 and a6=4a_6 = 4:

        • Common difference d=4−(−8)6−3=123=4d = \frac{4 - (-8)}{6 - 3} = \frac{12}{3} = 4

        • Zeroth term a0=−8−3(4)=−20a_0 = -8 - 3(4) = -20

        • Explicit rule: an=4n−20a_n = 4n - 20

  • Geometric Sequences:

    • Characteristics: Constant ratio / constant proportional change r=anan−1r = \frac{a_n}{a_{n-1}}.

    • Explicit Formula: an=a1rn−1a_n = a_1 r^{n-1}

    • Example 1: 2,6,18,54,…2, 6, 18, 54, \dots

      • Common ratio r=3r = 3

      • Seventh term a7=2(3)6=1458a_7 = 2(3)^6 = 1458

      • Explicit rule: an=2(3)n−1a_n = 2(3)^{n-1}

    • Example 2: 3,6,12,24,…3, 6, 12, 24, \dots

      • Common ratio r=2r = 2

      • Seventh term a7=3(2)6=192a_7 = 3(2)^6 = 192

      • Explicit rule: an=3(2)n−1a_n = 3(2)^{n-1}

    • Example 3: 1,−2,4,−8,16,…1, -2, 4, -8, 16, \dots

      • Common ratio r=−2r = -2

      • Seventh term a7=1(−2)6=64a_7 = 1(-2)^6 = 64

      • Explicit rule: an=1(−2)n−1a_n = 1(-2)^{n-1}

    • Example 4: −2,2,−2,2,…-2, 2, -2, 2, \dots

      • Common ratio r=−1r = -1

      • Seventh term a7=−2(−1)6=−2a_7 = -2(-1)^6 = -2

      • Explicit rule: an=−2(−1)n−1a_n = -2(-1)^{n-1}

    • Finding terms from given elements:

      • Geometric sequence with a3=3a_3 = 3 and a7=192a_7 = 192:

        • Common ratio relation: a7a3=r4=1923=64  ⟹  r=644=22\frac{a_7}{a_3} = r^4 = \frac{192}{3} = 64 \implies r = \sqrt[4]{64} = 2\sqrt{2}

        • Zeroth term a0=a3r3=3(22)3=3162=3232a_0 = \frac{a_3}{r^3} = \frac{3}{(2\sqrt{2})^3} = \frac{3}{16\sqrt{2}} = \frac{3\sqrt{2}}{32}

        • Explicit rule: an=a0rn=3232(22)na_n = a_0 r^n = \frac{3\sqrt{2}}{32}(2\sqrt{2})^n

  • Sequence Function Classification from Data Tables:

    • Bungy-Gungy Tree Growth in Amazon Rain Forest:

      • Time x (weeks)∈{0,1,2,3,4,5}x\,(weeks) \in \{0, 1, 2, 3, 4, 5\}

      • Height y (cm)∈{700,702.3,704.6,706.9,709.2,711.5}y\,(cm) \in \{700, 702.3, 704.6, 706.9, 709.2, 711.5\}

      • Function Type: Linear (Arithmetic sequence behavior)

      • Equation: y=2.3x+700y = 2.3x + 700

    • Thorium-232 Radioactive Decay (Half-life = 14 Billion Years):

      • Half-lives x∈{0,1,2,3,4,5}x \in \{0, 1, 2, 3, 4, 5\}

      • Mass y (grams)∈{16,8,4,2,1,0.5}y\,(grams) \in \{16, 8, 4, 2, 1, 0.5\}

      • Function Type: Exponential (Geometric sequence behavior)

      • Equation: y=16(0.5)xy = 16(0.5)^x

Sequence Convergence, Divergence, and Infinite Series

  • Summation Notation and Sum Evaluations:

    • Finite Series: −7+(−1)+5+11+⋯+53-7 + (-1) + 5 + 11 + \dots + 53

      • Sequence rule an=6n−13a_n = 6n - 13

      • Number of terms nn: 6n−13=53  ⟹  n=116n - 13 = 53 \implies n = 11

      • Summation notation: ∑n=111(6n−13)\sum_{n=1}^{11} (6n - 13)

      • Calculated sum: 253253

    • Finite Series: 2+5+8+11+⋯+292 + 5 + 8 + 11 + \dots + 29

      • Sequence rule an=3n−1a_n = 3n - 1

      • Number of terms nn: 3n−1=29  ⟹  n=103n - 1 = 29 \implies n = 10

      • Summation notation: ∑n=110(3n−1)\sum_{n=1}^{10} (3n - 1)

      • Calculated sum: 155155

    • Finite Series (8 terms): 6−12+24−48+…6 - 12 + 24 - 48 + \dots

      • Sequence rule an=6(−2)n−1a_n = 6(-2)^{n-1}

      • Summation notation: ∑n=186(−2)n−1\sum_{n=1}^{8} 6(-2)^{n-1}

      • Calculated sum: −510-510

    • Finite Geometric Sequence (n=12n = 12): 4,−2,1,…4, -2, 1, \dots

      • Summation notation: ∑n=1124(−0.5)n−1\sum_{n=1}^{12} 4(-0.5)^{n-1}

      • 12th term a12=4(−0.5)11=−1512a_{12} = 4(-0.5)^{11} = -\frac{1}{512}

      • Calculated sum: S12=1365512S_{12} = \frac{1365}{512}

  • Convergence vs. Divergence Analysis of Infinite Series:

    • Infinite Geometric Series Convergence Rule: An infinite geometric series ∑n=1∞a1rn−1\sum_{n=1}^{\infty} a_1 r^{n-1} converges if and only if ∣r∣<1|r| < 1. The sum is given by S=a11−rS = \frac{a_1}{1-r}.

    • Series 1: ∑n=1∞2(0.5)n\sum_{n=1}^{\infty} 2(0.5)^n

      • First term a1=1a_1 = 1, ratio r=0.5r = 0.5

      • ∣r∣=0.5<1  ⟹  |r| = 0.5 < 1 \implies Converges

      • Sum: S=11−0.5=2S = \frac{1}{1 - 0.5} = 2

    • Series 2: −3+3−3+3−…-3 + 3 - 3 + 3 - \dots

      • Ratio r=−1r = -1

      • ∣r∣=1≥1  ⟹  |r| = 1 \ge 1 \implies Diverges

    • Series 3: ∑n=1∞(5−0.4n)\sum_{n=1}^{\infty} (5 - 0.4n)

      • Arithmetic series with d=−0.4d = -0.4

      • Terms do not approach zero   ⟹  \implies Diverges

    • Series 4: 0.3+0.03+0.003+0.0003+…0.3 + 0.03 + 0.003 + 0.0003 + \dots

      • First term a1=0.3a_1 = 0.3, ratio r=0.1r = 0.1

      • ∣r∣=0.1<1  ⟹  |r| = 0.1 < 1 \implies Converges

      • Sum: S=0.31−0.1=0.30.9=13S = \frac{0.3}{1 - 0.1} = \frac{0.3}{0.9} = \frac{1}{3}

    • Series 5: 80+60+45+33.75+…80 + 60 + 45 + 33.75 + \dots

      • First term a1=80a_1 = 80, ratio r=0.75r = 0.75

      • ∣r∣=0.75<1  ⟹  |r| = 0.75 < 1 \implies Converges

      • Summation notation: ∑n=1∞80(0.75)n−1\sum_{n=1}^{\infty} 80(0.75)^{n-1}

      • Sum: S=801−0.75=800.25=320S = \frac{80}{1 - 0.75} = \frac{80}{0.25} = 320

Residual Analysis and Model Evaluation

  • Definition and Formula for Residuals:

    • A residual measures the vertical deviation of an actual data point from a modeled prediction.

    • Formula: Residual=yi−y^i=Actual Value−Modeled Value\text{Residual} = y_i - \hat{y}_i = \text{Actual Value} - \text{Modeled Value}

  • Case Study 1: Vertical Motion Ball Height Analysis:

    • Quadratic Predicted Height Model: s(t)=−4.676t2+3.758t+1.045s(t) = -4.676t^2 + 3.758t + 1.045

    • Alternative Exponential Fit: y=1.5264(0.8064)ty = 1.5264(0.8064)^t with r2=0.9918r^2 = 0.9918

    • Complete Experimental Data Table:

      • t=0.0000 st = 0.0000\,s: Actual Height =1.03754 m= 1.03754\,m, Predicted =1.5264 m= 1.5264\,m, Residual =−0.4889 m= -0.4889\,m

      • t=0.0180 st = 0.0180\,s: Actual Height =1.40205 m= 1.40205\,m, Predicted =1.5205 m= 1.5205\,m, Residual =−0.1184 m= -0.1184\,m

      • t=0.2150 st = 0.2150\,s: Actual Height =1.63806 m= 1.63806\,m, Predicted =1.4574 m= 1.4574\,m, Residual =0.1807 m= 0.1807\,m

      • t=0.3225 st = 0.3225\,s: Actual Height =1.77412 m= 1.77412\,m, Predicted =1.4241 m= 1.4241\,m, Residual =0.3501 m= 0.3501\,m

      • t=0.4300 st = 0.4300\,s: Actual Height =1.80392 m= 1.80392\,m, Predicted =1.3915 m= 1.3915\,m, Residual =0.4124 m= 0.4124\,m

      • t=0.5375 st = 0.5375\,s: Actual Height =1.71522 m= 1.71522\,m, Predicted =1.3597 m= 1.3597\,m, Residual =0.3555 m= 0.3555\,m

      • t=0.6450 st = 0.6450\,s: Actual Height =1.50942 m= 1.50942\,m, Predicted =1.3286 m= 1.3286\,m, Residual =0.1808 m= 0.1808\,m

      • t=0.7525 st = 0.7525\,s: Actual Height =1.21410 m= 1.21410\,m, Predicted =1.2982 m= 1.2982\,m, Residual =−0.0841 m= -0.0841\,m

      • t=0.8600 st = 0.8600\,s: Actual Height =0.83173 m= 0.83173\,m, Predicted =1.2685 m= 1.2685\,m, Residual =−0.4368 m= -0.4368\,m

    • Residual Plot Axes Window Settings: Domain [−0.1,1.0][-0.1, 1.0], Range [−0.02,0.02][-0.02, 0.02]

    • Model Evaluation Conclusion: A distinct curved pattern in the residual plot confirms that an exponential regression model is NOT a good fit for quadratic trajectory data.

  • Case Study 2: Exponential Model Residual Evaluation:

    • Fitted Model: f(x)=2.2652(0.7958)xf(x) = 2.2652(0.7958)^x

    • Correlation Parameters: r=−0.9959r = -0.9959, r2=0.9918r^2 = 0.9918

    • Complete Data and Residual Table:

      • x=−2x = -2: Actual f(x)=3.7f(x) = 3.7, Modeled =3.5768= 3.5768, Residual =0.1232= 0.1232

      • x=−1x = -1: Actual f(x)=2.7f(x) = 2.7, Modeled =2.8464= 2.8464, Residual =−0.1464= -0.1464

      • x=0x = 0: Actual f(x)=2.3f(x) = 2.3, Modeled =2.2652= 2.2652, Residual =0.0348= 0.0348

      • x=1x = 1: Actual f(x)=1.79f(x) = 1.79, Modeled =1.8026= 1.8026, Residual =−0.0126= -0.0126

      • x=2x = 2: Actual f(x)=1.45f(x) = 1.45, Modeled =1.4345= 1.4345, Residual =0.0155= 0.0155

AP Test Preparation and Practice Questions

  • Practice Question 1 (Arithmetic Sequence):

    • Problem: The first two terms of an arithmetic sequence are 22 and 88. What is the fourth term?

    • Derivation: Common difference d=8−2=6d = 8 - 2 = 6. Fourth term a4=2+3(6)=20a_4 = 2 + 3(6) = 20

    • Multiple Choice Options: a. 2020, b. 2626, c. 6464

    • Correct Answer: a. 20

  • Practice Question 2 (Geometric Sequence):

    • Problem: A geometric sequence begins with 2,6,…2, 6, \dots. What is the 5th term?

    • Derivation: Common ratio r=62=3r = \frac{6}{2} = 3. Fifth term a5=2(3)4=162a_5 = 2(3)^4 = 162

  • Practice Question 3 (Exponential Growth Rate):

    • Problem: What is the constant percentage growth rate of P(t)=1.23(1.049)xP(t) = 1.23(1.049)^x?

    • Derivation: Rate r=1.049−1=0.049=4.9%r = 1.049 - 1 = 0.049 = 4.9\%

    • Multiple Choice Options: a. 49%49\%, b. 23%23\%, c. 4.9%4.9\%, d. 2.3%2.3\%

    • Correct Answer: c. 4.9%

  • Practice Question 4 (Exponential Decay Rate):

    • Problem: What is the constant percentage decay rate of P(t)=22.7(0.834)xP(t) = 22.7(0.834)^x?

    • Derivation: Decay rate r=1−0.834=0.166=16.6%r = 1 - 0.834 = 0.166 = 16.6\%

    • Multiple Choice Options: a. 22.7%22.7\%, b. 16.6%16.6\%, c. 8.34%8.34\%, d. 2.27%2.27\%

    • Correct Answer: b. 16.6%

  • Practice Question 5 (Cell Division / Population Growth):

    • Problem: A single-cell amoeba divides into two every 4 days4\,days. About how long will it take one amoeba to produce a population of 10001000?

    • Derivation: 1000=1⋅2t/4  ⟹  t4=log⁡2(1000)≈9.9658  ⟹  t≈39.86 days1000 = 1 \cdot 2^{t/4} \implies \frac{t}{4} = \log_2(1000) \approx 9.9658 \implies t \approx 39.86\,days

    • Multiple Choice Options: a. 10 days10\,days, b. 20 days20\,days, c. 30 days30\,days, d. 40 days40\,days

    • Correct Answer: d. 40 days