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Mathematics · 3. Functions and graphs

Function graphs and transformations

Predict parameter effects, then test them using graphs, tables, and monotonic characteristic functions.

40 minutesContent version: 2.1
Start the interactive pathwayOpen the virtual experiment

Driving question

How can a formula change predict graph transformations and how can monotonicity solve complex equations?

Curriculum coverage

Vietnam Mathematics 9–10; Cambridge IGCSE/AS functions, graph transformations, exponential and logarithms.

Model scope and limitations

A screen graph shows a finite window and cannot alone prove global behaviour.

Learning objectives

  • Connect formulas, tables and graphs.
  • Recognise translations, stretches, reflections and asymptotes.
  • Apply characteristic monotonic functions and exponential-logarithmic equations.

Prerequisite knowledge

  • Cartesian coordinates and substitution.
  • Order of operations and monotonicity.

Core knowledge

Translations

y = f(x) + k shifts upward by k. y = f(x − h) shifts right by h.

Oblique asymptotes and symmetry center

Rational functions y = (ax² + bx + c)/(dx + e) have vertical asymptote x = −e/d and oblique asymptote y = αx + β via polynomial division. The intersection of both asymptotes is the symmetry center I.

Monotonic characteristic function method

If f(t) is continuous and strictly monotonic (f′(t) > 0 or f′(t) < 0) on interval K, then f(u) = f(v) is equivalent to u = v. This solves advanced polynomial, radical, exponential and logarithmic equations and systems.

Exponential equation classifications & methods

Four canonical methods for exponential equations: (1) Same base a^(f(x)) = a^(g(x)) ⇔ f(x) = g(x); (2) Substitution t = a^x > 0 for quadratic forms, homogeneous forms divided by b^(2x), or conjugate reciprocal bases (t + 1/t = m); (3) Logarithmization on both sides across differing bases; (4) Monotonic characteristic functions.

Logarithmic equation classifications & domain rules

Four canonical methods for logarithmic equations: (1) Same base logₐ(f(x)) = logₐ(g(x)) ⇔ f(x) = g(x) > 0; (2) Substitution t = logₐ(x) with change of base; (3) Exponentiation logₐ(f(x)) = g(x) ⇔ f(x) = a^(g(x)); (4) Monotonic functions. Traps: omitting domain f(x) > 0, not reversing inequality when 0 < a < 1, and omitting absolute values in logₐ(x²) = 2·logₐ(|x|).

Financial mathematics & Interest models

1. Simple: P(1+nr); 2. Compound: P(1+r)ⁿ (Rule of 72); 3. Continuous: P·e^(rt); 4. Annuity saving: [A(1+r)/r]·[(1+r)ⁿ − 1]; 5. Amortization loan: [P·r(1+r)ⁿ]/[(1+r)ⁿ − 1]; 6. Retirement depleting fund; 7. Depreciation & Malthus exponential growth N(t) = N₀·e^(kt).

Graphical root solver & V-diagram composite stitching

The equation f(x) = m corresponds to the intersection of y = f(x) with horizontal line y = m. Graph transformations: |f(x)| reflects below-x upward; f(|x|) reflects right half across y-axis. V-diagram composite stitching solves f(u(x)) = m by tracking core u(x) extrema through parent f(t) landmarks for rapid root counting without composite differentiation.

Worked example

Solve 2^(x+1) + 2^(x−1) = 10. Factor out 2^x: 2^x·(2 + 1/2) = 10 ⇔ 2^x = 4 ⇔ x = 2.

Misconceptions to avoid

  • The sign in f(x−h) produces a right shift by h, opposite to a naïve sign reading.
  • A finite graph window cannot prove the number of roots or global behaviour.
  • When solving exponential or logarithmic inequalities, a base 0 < a < 1 requires reversing the inequality sign.
  • Applying f(u) = f(v) ⇒ u = v is only valid when f(t) is strictly monotonic across the ENTIRE domain containing u and v.

Virtual experiment procedure: Function graphs explorer

  1. Choose a parent function such as f(x) = x² or sin x.
  2. Change one parameter at a time and predict before plotting.
  3. Record at least three key points and describe the transformation.

Safety and cautions

  • The viewing window may hide roots or asymptotes; always check domain and axis scales.

Evidence to collect

  • A table of at least three key points before and after transformation.
  • A transformation description verified by substitution, not appearance alone.
Open the virtual experiment
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<iframe src="https://visuallab.win/math-lab/?module=quadratic_func" width="100%" height="560" frameborder="0" allow="fullscreen" loading="lazy"></iframe>

Diagnostic check

1. How does y = f(x) + 3 compare with y = f(x)?

  1. Shift up 3
  2. Shift right 3
  3. Horizontal compression by 3
Show answer and explanation

Correct answer: Shift up 3

A constant added outside the function changes every y-value.

2. Where do solutions of f(x) = 0 appear on a graph?

  1. x-axis intersections
  2. y-axis intersection
  3. Only maxima
Show answer and explanation

Correct answer: x-axis intersections

On the x-axis, the y-coordinate is zero.

3. What is the domain condition of log₂(3x − 6) = 3?

  1. x > 2
  2. x ≥ 2
  3. x > 0
Show answer and explanation

Correct answer: x > 2

Argument inside logarithm must be strictly positive: 3x − 6 > 0 ⇔ x > 2.

Application practice

1. Where is the vertex of y = (x − 2)² + 1?

  1. (2, 1)
  2. (-2, 1)
  3. (1, 2)
Show answer and explanation

Correct answer: (2, 1)

The form (x − h)² + k has vertex (h, k) = (2, 1).

2. Which line is the oblique asymptote of y = (x² + 2x + 3)/(x + 1)?

  1. y = x + 1
  2. y = x
  3. y = 2x + 1
Show answer and explanation

Correct answer: y = x + 1

Polynomial division gives (x² + 2x + 3)/(x + 1) = x + 1 + 2/(x + 1), so the oblique asymptote is y = x + 1.

3. What is the solution of the exponential equation 2^(x+1) + 2^(x−1) = 10?

  1. x = 2
  2. x = 1
  3. x = 3
Show answer and explanation

Correct answer: x = 2

2^(x+1) + 2^(x−1) = 2^x·(5/2) = 10 ⇔ 2^x = 4 ⇔ x = 2.

4. What is the solution to log₃(2x − 1) = 2?

  1. x = 5
  2. x = 4
  3. x = 7
Show answer and explanation

Correct answer: x = 5

Condition 2x − 1 > 0 ⇔ x > 1/2. Exponentiating: 2x − 1 = 3² = 9 ⇔ x = 5 (valid).

5. What is the solution set of 4^x − 5·2^x + 4 = 0?

  1. {0, 2}
  2. {1, 4}
  3. {0, 1}
Show answer and explanation

Correct answer: {0, 2}

Let t = 2^x > 0: t² − 5t + 4 = 0 ⇔ t = 1 or t = 4. Hence 2^x = 1 ⇒ x = 0 and 2^x = 4 ⇒ x = 2.

6. A person deposits 100 million VND at 6% annual compound interest. How much is the total amount after 3 years?

  1. Approx 119.10 million VND
  2. 118.00 million VND
  3. 125.50 million VND
Show answer and explanation

Correct answer: Approx 119.10 million VND

Using compound interest: S_n = P(1 + r)ⁿ = 100 · (1.06)³ ≈ 119.10 million.

7. A cubic function has local maximum at y = 2 and local minimum at y = −2. How many real roots does f(x) = 1 have?

  1. 3 distinct roots
  2. 1 root
  3. 2 roots
Show answer and explanation

Correct answer: 3 distinct roots

Since the horizontal line y = 1 lies strictly between local minimum and maximum (-2 < 1 < 2), it intersects the cubic curve at exactly 3 distinct points.

Academic references

  1. Cambridge IGCSE Mathematics 0580 — official syllabus and resources
  2. Cambridge AS & A Level Mathematics 9709 — official syllabus
  3. OpenStax Precalculus 2e — functions, geometry and probability

This lesson is maintained against the cited sources below. Simulations are learning models and do not replace supervised physical-laboratory safety procedures.

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