High School Mathematics

Grade 11: Algebra 2 & Trigonometry

Master advanced algebra, trigonometric functions, complex numbers, and analytical reasoning with comprehensive lessons and authentic practice exams

Practice Worksheets Available

Reinforce your Algebra II skills with our comprehensive worksheet collection

Welcome to Algebra 2 & Trigonometry!

This year, you'll explore the beauty of advanced functions, master trigonometric identities, work with complex numbers, and develop sophisticated problem-solving skills. Each lesson includes detailed explanations, worked examples, and 30 practice questions with complete step-by-step solutions.

— Mr. Augustine, Mathematics Department

Polynomial Functions

Master polynomial operations, end behavior, zeros, and graphing techniques

Key Vocabulary

PolynomialDegreeLeading CoefficientEnd BehaviorZeros/RootsMultiplicityTurning PointsSynthetic DivisionRemainder TheoremFactor Theorem

Concept Explanation

A polynomial function is a function of the form f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where n is a non-negative integer and the coefficients are real numbers. **Key Properties:** • The degree of a polynomial is the highest power of x with a non-zero coefficient • The leading coefficient is the coefficient of the term with the highest degree • End behavior describes what happens to f(x) as x approaches positive or negative infinity • A polynomial of degree n has at most n real zeros and at most n-1 turning points **End Behavior Rules:** • If degree is even and leading coefficient is positive: both ends go up • If degree is even and leading coefficient is negative: both ends go down • If degree is odd and leading coefficient is positive: left down, right up • If degree is odd and leading coefficient is negative: left up, right down **Finding Zeros:** • Factor the polynomial completely • Set each factor equal to zero • The multiplicity of a zero determines the behavior at that x-intercept: - Odd multiplicity: graph crosses the x-axis - Even multiplicity: graph touches the x-axis and turns around **Synthetic Division:** Used to divide a polynomial by (x - c). The remainder equals f(c) by the Remainder Theorem.

Worked Examples

1

Find the end behavior of f(x) = -2x⁵ + 3x³ - x + 7

Step 1: Identify the degree and leading coefficient Degree = 5 (odd) Leading coefficient = -2 (negative) Step 2: Apply end behavior rules Since degree is odd and leading coefficient is negative: • As x → -∞, f(x) → +∞ (left side goes up) • As x → +∞, f(x) → -∞ (right side goes down) Answer: The graph rises to the left and falls to the right.
2

Find all zeros and their multiplicities: f(x) = x⁴ - 5x³ + 4x²

Step 1: Factor completely f(x) = x⁴ - 5x³ + 4x² f(x) = x²(x² - 5x + 4) f(x) = x²(x - 4)(x - 1) Step 2: Set each factor equal to zero x² = 0 → x = 0 (multiplicity 2) x - 4 = 0 → x = 4 (multiplicity 1) x - 1 = 0 → x = 1 (multiplicity 1) Step 3: Describe behavior at each zero • At x = 0: graph touches and turns (even multiplicity) • At x = 1: graph crosses (odd multiplicity) • At x = 4: graph crosses (odd multiplicity) Answer: Zeros are x = 0 (mult. 2), x = 1 (mult. 1), x = 4 (mult. 1)
3

Use synthetic division to divide (2x³ - 5x² + 3x - 8) by (x - 2)

Step 1: Set up synthetic division with c = 2 2 | 2 -5 3 -8 | 4 -2 2 ________________ 2 -1 1 -6 Step 2: Interpret the result Quotient: 2x² - x + 1 Remainder: -6 Step 3: Write the result (2x³ - 5x² + 3x - 8) ÷ (x - 2) = 2x² - x + 1 - 6/(x-2) By the Remainder Theorem: f(2) = -6 Answer: Quotient is 2x² - x + 1, remainder is -6

Mr. Augustine's Teaching Tip

When graphing polynomials, always start by finding the zeros and checking end behavior. Remember: the graph of a polynomial is always smooth and continuous—no breaks, holes, or sharp corners! If you're stuck factoring, try grouping or use the Rational Root Theorem to test possible rational zeros.

Rational Functions

Explore asymptotes, holes, domain restrictions, and graphing rational functions

Key Vocabulary

Rational FunctionVertical AsymptoteHorizontal AsymptoteSlant/Oblique AsymptoteHole/Removable DiscontinuityDomainRangeEnd BehaviorIntercepts

Concept Explanation

A rational function is a function of the form f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials and Q(x) ≠ 0. **Domain:** All real numbers except values that make the denominator zero. **Vertical Asymptotes:** Occur at values of x that make the denominator zero but NOT the numerator (after simplification). The graph approaches ±∞ near vertical asymptotes. **Holes (Removable Discontinuities):** Occur when a factor cancels from both numerator and denominator. The graph has a "hole" at that x-value. **Horizontal Asymptotes:** Determined by comparing degrees of numerator (n) and denominator (m): • If n < m: horizontal asymptote at y = 0 • If n = m: horizontal asymptote at y = (leading coefficient of P)/(leading coefficient of Q) • If n > m: no horizontal asymptote (may have slant asymptote) **Slant (Oblique) Asymptotes:** Occur when degree of numerator is exactly 1 more than degree of denominator. Find by polynomial long division. **Graphing Strategy:** 1. Factor numerator and denominator 2. Find domain (exclude zeros of denominator) 3. Identify holes (common factors) 4. Find vertical asymptotes (remaining zeros of denominator) 5. Find horizontal or slant asymptote 6. Find intercepts 7. Plot points and sketch

Worked Examples

1

Find all asymptotes and holes for f(x) = (x² - 4)/(x² - x - 6)

Step 1: Factor numerator and denominator f(x) = (x - 2)(x + 2) / [(x - 3)(x + 2)] Step 2: Identify holes Common factor: (x + 2) Hole at x = -2 To find y-coordinate: simplify first, then substitute f(x) = (x - 2)/(x - 3) for x ≠ -2 f(-2) = (-2 - 2)/(-2 - 3) = -4/-5 = 4/5 Hole at (-2, 4/5) Step 3: Find vertical asymptotes After canceling, denominator is (x - 3) Vertical asymptote: x = 3 Step 4: Find horizontal asymptote Degrees are equal (both 2 in original form, both 1 after simplification) y = (leading coefficient of numerator)/(leading coefficient of denominator) = 1/1 = 1 Horizontal asymptote: y = 1 Answer: Hole at (-2, 4/5), vertical asymptote x = 3, horizontal asymptote y = 1
2

Find the slant asymptote of f(x) = (2x² + 3x - 1)/(x - 2)

Step 1: Check if slant asymptote exists Degree of numerator (2) is exactly 1 more than degree of denominator (1) Yes, slant asymptote exists Step 2: Use polynomial long division 2x + 7 ________ x - 2 | 2x² + 3x - 1 2x² - 4x ________ 7x - 1 7x - 14 _______ 13 Step 3: The quotient is the slant asymptote Slant asymptote: y = 2x + 7 Answer: y = 2x + 7
3

Graph f(x) = (x + 1)/(x - 2) by finding all key features

Step 1: Domain x ≠ 2 (denominator cannot be zero) Step 2: Vertical asymptote x = 2 (denominator is zero, numerator is not) Step 3: Horizontal asymptote Degrees are equal (both 1) y = 1/1 = 1 Step 4: x-intercept Set numerator = 0: x + 1 = 0 → x = -1 x-intercept: (-1, 0) Step 5: y-intercept f(0) = (0 + 1)/(0 - 2) = 1/(-2) = -1/2 y-intercept: (0, -1/2) Step 6: Behavior near vertical asymptote Test x = 1.9: f(1.9) = 2.9/(-0.1) = -29 (goes to -∞ from left) Test x = 2.1: f(2.1) = 3.1/0.1 = 31 (goes to +∞ from right) Step 7: Sketch Graph has two branches separated by vertical asymptote at x = 2 Both branches approach horizontal asymptote y = 1 as x → ±∞

Mr. Augustine's Teaching Tip

Always factor first! Many students miss holes because they don't simplify before finding asymptotes. Remember: holes occur where factors cancel, vertical asymptotes occur where the denominator is zero after simplification. When graphing, plot the asymptotes first—they're your 'boundaries'—then plot intercepts and a few key points to see the shape.

Radical Functions

Master square roots, cube roots, simplifying radicals, and solving radical equations

Key Vocabulary

RadicalRadicandIndexSquare RootCube RootRational ExponentSimplifyRationalizeExtraneous SolutionDomain Restriction

Concept Explanation

A radical function contains a variable under a radical symbol. The most common are square root and cube root functions. **Radical Notation:** ⁿ√x where n is the index and x is the radicand • √x means ²√x (square root) • ³√x is cube root • ⁿ√x = x^(1/n) **Properties of Radicals:** • √(ab) = √a · √b • √(a/b) = √a / √b • (√a)² = a (for a ≥ 0) • ⁿ√(aⁿ) = |a| if n is even, = a if n is odd **Simplifying Radicals:** 1. Factor the radicand into perfect powers 2. Take roots of perfect powers 3. Simplify remaining factors **Rationalizing Denominators:** • For √a in denominator: multiply by √a/√a • For (a + √b) in denominator: multiply by conjugate (a - √b)/(a - √b) **Domain of Square Root Functions:** For f(x) = √g(x), domain is all x where g(x) ≥ 0 **Solving Radical Equations:** 1. Isolate the radical 2. Raise both sides to the power equal to the index 3. Solve the resulting equation 4. CHECK all solutions (extraneous solutions may occur!) **Rational Exponents:** • x^(m/n) = ⁿ√(x^m) = (ⁿ√x)^m • x^(-m/n) = 1/x^(m/n)

Worked Examples

1

Simplify: √72

Step 1: Factor 72 into perfect squares 72 = 36 × 2 = 6² × 2 Step 2: Apply the product property √72 = √(36 × 2) = √36 · √2 Step 3: Simplify √72 = 6√2 Answer: 6√2
2

Solve: √(2x + 3) = 5

Step 1: Square both sides [√(2x + 3)]² = 5² 2x + 3 = 25 Step 2: Solve for x 2x = 22 x = 11 Step 3: CHECK the solution √(2(11) + 3) = √(22 + 3) = √25 = 5 ✓ Answer: x = 11
3

Rationalize the denominator: 3/(2 + √5)

Step 1: Multiply by the conjugate 3/(2 + √5) · (2 - √5)/(2 - √5) Step 2: Multiply numerators and denominators Numerator: 3(2 - √5) = 6 - 3√5 Denominator: (2 + √5)(2 - √5) = 4 - 5 = -1 Step 3: Simplify (6 - 3√5)/(-1) = -6 + 3√5 = 3√5 - 6 Answer: 3√5 - 6 or -6 + 3√5

Mr. Augustine's Teaching Tip

Always CHECK your solutions when solving radical equations! When you square both sides, you can introduce extraneous solutions that don't actually work in the original equation. Also, remember that even-indexed radicals (like square roots) require non-negative radicands, which affects the domain. When simplifying, look for the largest perfect square factor—it saves time!

Exponential & Logarithmic Functions

Explore exponential growth/decay, logarithms, properties, and solving exponential equations

Key Vocabulary

Exponential FunctionBaseGrowthDecayLogarithmCommon LogNatural LogChange of BaseCompound InterestHalf-life

Concept Explanation

**Exponential Functions:** f(x) = a · b^x where a ≠ 0, b > 0, b ≠ 1 • If b > 1: exponential growth • If 0 < b < 1: exponential decay • Domain: all real numbers • Range: y > 0 (if a > 0) • Horizontal asymptote: y = 0 **Special Exponential Function:** f(x) = e^x where e ≈ 2.71828 (Euler's number) **Logarithmic Functions:** y = log_b(x) means b^y = x • log_b(x) is the inverse of b^x • Domain: x > 0 • Range: all real numbers • Vertical asymptote: x = 0 **Special Logarithms:** • Common log: log(x) = log₁₀(x) • Natural log: ln(x) = log_e(x) **Properties of Logarithms:** • log_b(xy) = log_b(x) + log_b(y) (Product Rule) • log_b(x/y) = log_b(x) - log_b(y) (Quotient Rule) • log_b(x^n) = n · log_b(x) (Power Rule) • log_b(b) = 1 • log_b(1) = 0 • b^(log_b(x)) = x • log_b(b^x) = x **Change of Base Formula:** log_b(x) = log(x)/log(b) = ln(x)/ln(b) **Solving Exponential Equations:** 1. If bases are the same: set exponents equal 2. If bases differ: take log of both sides **Applications:** • Compound Interest: A = P(1 + r/n)^(nt) • Continuous Growth: A = Pe^(rt) • Half-life: A = A₀(1/2)^(t/h)

Worked Examples

1

Solve: 2^(x+1) = 32

Step 1: Express 32 as a power of 2 32 = 2^5 Step 2: Set exponents equal 2^(x+1) = 2^5 x + 1 = 5 Step 3: Solve for x x = 4 Answer: x = 4
2

Solve: 5^x = 20 (use logarithms)

Step 1: Take log of both sides log(5^x) = log(20) Step 2: Use power rule x · log(5) = log(20) Step 3: Solve for x x = log(20)/log(5) x ≈ 1.861 Answer: x = log(20)/log(5) ≈ 1.861
3

Expand: log₃(x²y/z)

Step 1: Apply quotient rule log₃(x²y/z) = log₃(x²y) - log₃(z) Step 2: Apply product rule to numerator log₃(x²y) = log₃(x²) + log₃(y) Step 3: Apply power rule log₃(x²) = 2log₃(x) Step 4: Combine log₃(x²y/z) = 2log₃(x) + log₃(y) - log₃(z) Answer: 2log₃(x) + log₃(y) - log₃(z)

Mr. Augustine's Teaching Tip

Logarithms are just exponents in disguise! When you see log₂(8) = 3, think '2 to what power equals 8? Answer: 3.' This connection makes everything clearer. For solving exponential equations, if you can't make the bases match, take the log of both sides—it's your secret weapon. And remember: you can only take the log of positive numbers, so always check your domain!

Unit Circle & Radian Measure

Master the unit circle, convert between degrees and radians, and find exact trig values

Key Vocabulary

Unit CircleRadianDegreeArc LengthReference AngleQuadrantTerminal SideStandard PositionCoterminal Angles

Concept Explanation

**The Unit Circle:** A circle with radius 1 centered at the origin. Used to define trigonometric functions for all angles. **Radian Measure:** One radian is the angle formed when the arc length equals the radius. • Full circle: 2π radians = 360° • Half circle: π radians = 180° • Conversion: radians = degrees × (π/180) • Conversion: degrees = radians × (180/π) **Key Angle Conversions:** • 30° = π/6 rad • 45° = π/4 rad • 60° = π/3 rad • 90° = π/2 rad • 180° = π rad • 270° = 3π/2 rad • 360° = 2π rad **Unit Circle Coordinates:** For angle θ in standard position, the point on the unit circle is (cos θ, sin θ) **Special Angles (Exact Values):** • 0°: (1, 0) → cos = 1, sin = 0 • 30° (π/6): (√3/2, 1/2) • 45° (π/4): (√2/2, √2/2) • 60° (π/3): (1/2, √3/2) • 90° (π/2): (0, 1) → cos = 0, sin = 1 **Reference Angles:** The acute angle formed with the x-axis. Used to find trig values in all quadrants. **Quadrant Signs:** • QI: All positive • QII: Sin positive, Cos/Tan negative • QIII: Tan positive, Sin/Cos negative • QIV: Cos positive, Sin/Tan negative (Remember: "All Students Take Calculus") **Coterminal Angles:** Angles that share the same terminal side. Add or subtract 360° (or 2π radians).

Worked Examples

1

Convert 135° to radians

Step 1: Use conversion formula radians = degrees × (π/180) Step 2: Substitute and simplify radians = 135 × (π/180) radians = 135π/180 Step 3: Reduce the fraction radians = 3π/4 Answer: 3π/4 radians
2

Find the exact value of sin(5π/6)

Step 1: Identify the quadrant 5π/6 is in Quadrant II (between π/2 and π) Step 2: Find the reference angle Reference angle = π - 5π/6 = π/6 Step 3: Find sin(π/6) sin(π/6) = 1/2 Step 4: Apply quadrant sign In QII, sine is positive sin(5π/6) = 1/2 Answer: 1/2
3

Find a positive and negative coterminal angle for 50°

Step 1: Add 360° for positive coterminal angle 50° + 360° = 410° Step 2: Subtract 360° for negative coterminal angle 50° - 360° = -310° Answer: Positive: 410°, Negative: -310°

Mr. Augustine's Teaching Tip

The unit circle is your best friend in trigonometry! Memorize the coordinates for 0°, 30°, 45°, 60°, and 90° in the first quadrant—then use reference angles and the 'All Students Take Calculus' rule to find values in other quadrants. For radians, think of π as 180°, so π/2 is half of that (90°), π/4 is a quarter (45°), etc. Draw the unit circle whenever you're stuck!

Trigonometric Functions & Graphs

Graph sine, cosine, and tangent functions with transformations, amplitude, period, and phase shift

Key Vocabulary

AmplitudePeriodPhase ShiftVertical ShiftMidlineFrequencySinusoidalAsymptoteMaximumMinimum

Concept Explanation

**General Form of Sinusoidal Functions:** f(x) = A sin[B(x - C)] + D or f(x) = A cos[B(x - C)] + D Where: • |A| = amplitude (distance from midline to max/min) • Period = 2π/|B| • C = phase shift (horizontal shift) • D = vertical shift (midline) **Parent Functions:** • y = sin(x): starts at (0,0), period 2π, amplitude 1 • y = cos(x): starts at (0,1), period 2π, amplitude 1 • y = tan(x): period π, vertical asymptotes at x = π/2 + nπ **Key Features of y = sin(x):** • Domain: all real numbers • Range: [-1, 1] • Period: 2π • Zeros: x = nπ (where n is an integer) • Maximum: 1 at x = π/2 + 2nπ • Minimum: -1 at x = 3π/2 + 2nπ **Key Features of y = cos(x):** • Domain: all real numbers • Range: [-1, 1] • Period: 2π • Zeros: x = π/2 + nπ • Maximum: 1 at x = 2nπ • Minimum: -1 at x = π + 2nπ **Key Features of y = tan(x):** • Domain: all real numbers except x = π/2 + nπ • Range: all real numbers • Period: π • Zeros: x = nπ • Vertical asymptotes: x = π/2 + nπ **Transformations:** • A < 0: reflection over x-axis • B > 1: horizontal compression (shorter period) • 0 < B < 1: horizontal stretch (longer period) • C > 0: shift right • C < 0: shift left • D > 0: shift up • D < 0: shift down

Worked Examples

1

Find the amplitude, period, and phase shift of f(x) = 3sin(2x - π) + 1

Step 1: Rewrite in standard form f(x) = 3sin[2(x - π/2)] + 1 Step 2: Identify A, B, C, D A = 3, B = 2, C = π/2, D = 1 Step 3: Calculate features Amplitude = |A| = |3| = 3 Period = 2π/|B| = 2π/2 = π Phase shift = C = π/2 (right) Vertical shift = D = 1 (up) Midline: y = 1 Answer: Amplitude = 3, Period = π, Phase shift = π/2 right, Vertical shift = 1 up
2

Graph y = -2cos(x) + 1

Step 1: Identify transformations A = -2 (amplitude 2, reflected over x-axis) B = 1 (period = 2π/1 = 2π) C = 0 (no phase shift) D = 1 (shift up 1) Step 2: Find key features Midline: y = 1 Maximum: y = 1 + 2 = 3 (at x = π, 3π, ...) Minimum: y = 1 - 2 = -1 (at x = 0, 2π, ...) Step 3: Plot key points (0, -1), (π/2, 1), (π, 3), (3π/2, 1), (2π, -1) Step 4: Sketch smooth curve through points The graph is a cosine curve reflected over the x-axis, with amplitude 2 and midline y = 1.
3

Find the equation of a sine function with amplitude 4, period π, and phase shift π/4 right

Step 1: Use general form f(x) = A sin[B(x - C)] + D Step 2: Find A from amplitude Amplitude = 4, so A = 4 (or -4) Use A = 4 Step 3: Find B from period Period = 2π/B = π B = 2π/π = 2 Step 4: Find C from phase shift Phase shift = π/4 right, so C = π/4 Step 5: Assume no vertical shift D = 0 Answer: f(x) = 4sin[2(x - π/4)] or f(x) = 4sin(2x - π/2)

Mr. Augustine's Teaching Tip

When graphing trig functions, always start by finding the five key points for one period. For sine: start, max, middle, min, end. For cosine: max, middle, min, middle, max. The period tells you how wide one cycle is, and the amplitude tells you how tall. Remember: B affects the period (2π/B), not the amplitude! And always check if there's a negative sign in front—that flips the graph upside down.

Trigonometric Identities

Master Pythagorean, reciprocal, quotient, and cofunction identities to simplify and verify expressions

Key Vocabulary

IdentityPythagorean IdentityReciprocal IdentityQuotient IdentityCofunction IdentityEven/Odd IdentityVerifySimplifyFundamental Identities

Concept Explanation

**Fundamental Trigonometric Identities:** **Reciprocal Identities:** • csc(θ) = 1/sin(θ) • sec(θ) = 1/cos(θ) • cot(θ) = 1/tan(θ) **Quotient Identities:** • tan(θ) = sin(θ)/cos(θ) • cot(θ) = cos(θ)/sin(θ) **Pythagorean Identities:** • sin²(θ) + cos²(θ) = 1 • 1 + tan²(θ) = sec²(θ) • 1 + cot²(θ) = csc²(θ) **Cofunction Identities:** • sin(90° - θ) = cos(θ) • cos(90° - θ) = sin(θ) • tan(90° - θ) = cot(θ) **Even/Odd Identities:** • sin(-θ) = -sin(θ) (odd) • cos(-θ) = cos(θ) (even) • tan(-θ) = -tan(θ) (odd) **Strategies for Verifying Identities:** 1. Work with one side at a time (usually the more complex side) 2. Convert everything to sine and cosine 3. Use algebraic techniques (factor, combine fractions, multiply by conjugate) 4. Look for opportunities to use Pythagorean identities 5. Never move terms from one side to the other (not an equation to solve!) **Common Techniques:** • Factor when possible • Multiply by conjugate to eliminate radicals • Find common denominators • Substitute using fundamental identities • Split fractions

Worked Examples

1

Verify the identity: tan(θ) · cos(θ) = sin(θ)

Step 1: Work with the left side tan(θ) · cos(θ) Step 2: Convert tan to sin/cos = [sin(θ)/cos(θ)] · cos(θ) Step 3: Simplify = sin(θ) · [cos(θ)/cos(θ)] = sin(θ) · 1 = sin(θ) Step 4: This equals the right side ✓ Identity verified!
2

Simplify: (1 - sin²(θ))/cos(θ)

Step 1: Use Pythagorean identity 1 - sin²(θ) = cos²(θ) Step 2: Substitute = cos²(θ)/cos(θ) Step 3: Simplify = cos(θ) Answer: cos(θ)
3

Verify: (sin(θ) + cos(θ))² = 1 + 2sin(θ)cos(θ)

Step 1: Expand the left side (sin(θ) + cos(θ))² = sin²(θ) + 2sin(θ)cos(θ) + cos²(θ) Step 2: Rearrange = [sin²(θ) + cos²(θ)] + 2sin(θ)cos(θ) Step 3: Use Pythagorean identity = 1 + 2sin(θ)cos(θ) Step 4: This equals the right side ✓ Identity verified!

Mr. Augustine's Teaching Tip

When verifying identities, NEVER treat it like an equation you're solving—you can't add or subtract from both sides! Instead, work with one side (usually the more complicated one) and transform it step-by-step until it matches the other side. My favorite trick: when stuck, convert everything to sine and cosine. Also, memorize the three Pythagorean identities—they're your most powerful tools!

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