Flashcards · IB Diploma Programme Mathematics: Applications and Interpretation · Topic 2

Functions flashcards

51 cards on Topic 2, subtopics 2.1 to 2.10. Pick a subtopic or study them all.

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2.1 Different forms of the equation of a straight line; gradient, parallel and perpendicular lines

Three forms of the equation of a straight line: Gradient-intercept: y = mx + c General: ax + by + d = 0 (gradient −a/b) Point-gradient: y − y₁ = m(x − x₁)
Gradient from two points; parallel and perpendicular lines: m = (y₂ − y₁)/(x₂ − x₁). Parallel: m₁ = m₂. Perpendicular: m₁ × m₂ = −1.
Find the gradient of 4x − 2y + 7 = 0: 2y = 4x + 7, so y = 2x + 3.5 and the gradient is 2.
A line has equation y = −3x + 1. State the gradient of any line perpendicular to it: 1/3, the negative reciprocal of −3.
Find the line through (0, 3) and (4, 11) in the form y = mx + c: m = (11 − 3)/4 = 2 and c = 3, so y = 2x + 3.

2.2 Concept of a function, domain, range and graph; inverse function

Function, domain and range: Function: a rule giving exactly one output for each input. Domain: the set of inputs (in a model, also limited by the context). Range: the set of outputs, written in terms of f(x) or y.
Inverse function f⁻¹: when it exists and its graph: It exists only when f is one-to-one. Its graph is the reflection of y = f(x) in y = x, and the domain of f⁻¹ is the range of f.
How do you find an inverse function? Use f(x) = 4x − 7: 1. Write y = 4x − 7. 2. Make x the subject: x = (y + 7)/4. 3. Swap: f⁻¹(x) = (x + 7)/4. 4. State the domain of f⁻¹ (the range of f).
f⁻¹(x) vs 1/f(x): f⁻¹(x) is the inverse function, which undoes f. It is not the reciprocal 1/f(x).
State the domain and range of f(x) = 3 + √(x − 1): Domain x ≥ 1 (the square root needs x − 1 ≥ 0). Range f(x) ≥ 3.

2.3 The graph of a function; sketching from information or a context

Command terms: sketch vs draw: Sketch: the general shape with key features (intercepts, turning points, asymptotes) labelled in roughly the right places. Draw: plotted accurately and to scale, usually on given axes.
Copying a graph from your GDC to paper: what must you include?: Labelled axes, the viewing window you used marked on the axes, and every key feature labelled with its coordinates or equation.
How do you graph a profit function from revenue R(x) and cost C(x)?: Profit P(x) = R(x) − C(x). Graph y = R(x) − C(x) directly on the GDC. Example: R = 30x − 0.1x², C = 200 + 4x gives P = −0.1x² + 26x − 200.
You are asked to sketch a function not named in the syllabus. What do you do?: The same as for any function: graph it on the GDC over a suitable window, then sketch the shape and label its key features.

2.4 Determining key features of graphs; points of intersection of two curves using technology

How do you find the key features of a graph on a GDC?: 1. Choose a window that shows every feature. 2. Zero/root for x-intercepts; evaluate at x = 0 for the y-intercept. 3. Max/min for turning points (and the vertex). 4. Trace to large |x| and near excluded values for asymptotes.
How do you find where two curves meet using technology?: Graph y = f(x) and y = g(x), use intersect once for each point, and write ‘f(x) = g(x) solved on GDC’ with both coordinates. Widen the window so no point is missed.
Horizontal and vertical asymptotes, and how to write them: Horizontal: a line y = k approached as |x| gets large. Vertical: a line x = k where the graph grows without bound. Always write them as equations, such as y = 20 or x = 2.
Find the vertex of y = 2x² − 12x + 5: x = −b/(2a) = 12/4 = 3; y = 18 − 36 + 5 = −13. Vertex (3, −13).
Find where y = x² meets y = 2x + 3: x² − 2x − 3 = 0, so x = 3 or x = −1. Points (3, 9) and (−1, 1).

2.5 Modelling with linear, quadratic, exponential, direct/inverse variation, cubic and sinusoidal functions

Quadratic model f(x) = ax² + bx + c: key facts: a ≠ 0. Axis of symmetry x = −b/(2a), with the vertex on it. y-intercept c. One turning point.
Exponential model kaˣ + c: asymptote and value at x = 0: Horizontal asymptote y = c; value at x = 0 is k + c. With a > 1, kaˣ + c grows and ka⁻ˣ + c decays (for k > 0).
Direct and inverse variation model: f(x) = axⁿ, n an integer. n > 0: y grows as x grows. n < 0: y shrinks as x grows, and the y-axis is a vertical asymptote.
Sinusoidal model a sin(bx) + d (degrees): amplitude, period, principal axis: Amplitude |a|; period 360°/b; principal axis y = d; max d + |a|, min d − |a|. Example: h(t) = −12 cos(18t) + 15 has amplitude 12, period 20, max 27, min 3.
Which model fits: steady change, one turning point, constant factor, two turning points, repeating?: Steady increase or decrease: linear. One symmetric turning point: quadratic. Constant growth or decay factor, levelling off: exponential. Two turning points: cubic. Repeating pattern: sinusoidal.

2.6 Modelling skills: developing, fitting, testing and using a model

The three stages of modelling in the AI guide: 1. Develop and fit the model (choose a type, find its parameters, state a domain). 2. Test and reflect (is it appropriate and reasonable?). 3. Use it (read values, interpret, predict).
How do you fit a model's parameters at SL?: Use initial conditions first (the value at x = 0 often gives k or c), substitute other points to get equations, and solve them simultaneously on the GDC (up to three unknowns). Write the model in full and check one point.
N(t) = k·aᵗ with N(0) = 120 and N(5) = 300. Find k and a: k = 120. 120a⁵ = 300, so a⁵ = 2.5 and a = 2.5^(1/5) = 1.20 (3 s.f.). Keep the unrounded a for predictions.
Extrapolation, and why it is risky: Using a model outside the range of the data used to build it. It assumes the pattern continues, which it often does not.
How do you check data to choose between a linear and an exponential model?: Linear: constant differences between successive values. Exponential: constant ratios. Check the data both ways.

2.7 Composite functions in context; inverse function including domain restriction (AHL only)

Composite function (f ∘ g)(x)HL: (f ∘ g)(x) = f(g(x)): apply g first, then f. Example: f(x) = 2x + 1, g(x) = x² gives (f ∘ g)(3) = f(9) = 19, but (g ∘ f)(3) = g(7) = 49.
Inverse function: identity and domain/range swapHL: (f ∘ f⁻¹)(x) = (f⁻¹ ∘ f)(x) = x. Domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.
Domain restrictionHL: Limiting the domain of a many-to-one function so that it becomes one-to-one and has an inverse. Example: f(x) = (x − 2)² + 1, x ≥ 2 has f⁻¹(x) = 2 + √(x − 1), x ≥ 1.
Finding an inverse that needs a square root: which sign?HL: Choose the sign that matches the restricted domain of f. Do not leave ± in the answer.
A conversion c(d) = 0.92d is followed by a fee k(e) = 0.98e − 2. Find the single functionHL: (k ∘ c)(d) = 0.98(0.92d) − 2 = 0.9016d − 2. c is applied first, so it goes inside.

2.8 Transformations of graphs: translations, reflections, stretches and composite transformations (AHL only)

y = f(x − a) + b: what transformation?HL: Translation a units right and b units up, by the vector (a, b). Note y = f(x + 3) moves the graph 3 units left.
y = −f(x) and y = f(−x)HL: y = −f(x): reflection in the x-axis. y = f(−x): reflection in the y-axis.
y = p f(x) and y = f(qx)HL: y = p f(x): vertical stretch, scale factor p. y = f(qx): horizontal stretch, scale factor 1/q. So f(2x) is a horizontal stretch by 1/2.
Order of steps from y = f(x) to y = p f(q(x − a)) + bHL: 1. Horizontal stretch, factor 1/q. 2. Horizontal translation by a. 3. Vertical stretch, factor p. 4. Vertical translation by b. The vertical stretch must come before the vertical translation.
(4, 6) lies on y = f(x). Find its image on y = 3f(x + 2) − 1HL: x: 4 − 2 = 2. y: 3 × 6 − 1 = 17. Image (2, 17).

2.9 Further modelling: exponential (half-life), natural logarithmic, sinusoidal, logistic and piecewise models (AHL only)

Half-life modelHL: A = A₀e^(−kt) with k = ln 2 ÷ half-life, or A = A₀(0.5)^(t/half-life). Example: A = A₀e^(−0.0462t) has half-life ln 2/0.0462 = 15.0 (3 s.f.).
Logistic model: form and key featuresHL: f(x) = L/(1 + Ce^(−kx)) with L, C, k > 0. Horizontal asymptote y = L (the carrying capacity). f(0) = L/(1 + C), so C = L/f(0) − 1.
Sinusoidal model in radians: f(x) = a sin(b(x − c)) + dHL: Amplitude |a|, period 2π/b, phase shift c, principal axis y = d. HL models are in radians unless a degree sign appears.
How do you find a and d of a sinusoidal model from its maximum and minimum?HL: a = (max − min)/2 and d = (max + min)/2. Example: max 15, min 3 gives |a| = 6, d = 9.
Logarithmic model vs logistic modelHL: f(x) = a + b ln x (b > 0) keeps increasing without a bound, only more and more slowly. A logistic curve levels off towards y = L.
Piecewise model: f(x) = 3x for x < 2 and ax − 4 for x ≥ 2. Find a so that f is continuousHL: Make the pieces agree at x = 2: 3(2) = 2a − 4, so a = 5.

2.10 Scaling large or small numbers with logarithms; linearizing data; log-log and semi-log graphs (AHL only)

Semi-log and log-log graphs: what does a straight line show?HL: Semi-log (log y against x) straight: an exponential model y = kaˣ. Log-log (log y against log x) straight: a power model y = axᵇ.
Linearize y = kaˣ and y = axᵇHL: y = kaˣ ⇒ ln y = ln k + x ln a. y = axᵇ ⇒ ln y = ln a + b ln x.
How do you choose between an exponential and a power model by linearizing?HL: 1. Find r for (x, ln y) and for (ln x, ln y). 2. The pair with |r| closer to 1 is the better linear fit. 3. Regress that pair, then convert: exponential k = e^(intercept), a = e^(gradient); power a = e^(intercept), b = gradient.
Data satisfy log₁₀ y = 0.25x + 0.9. Write y in the form kaˣHL: k = 10^0.9 = 7.94 and a = 10^0.25 = 1.78, so y = 7.94 × 1.78ˣ.
Data satisfy ln y = 0.7 ln x + 1.2. Write y in the form axᵇHL: a = e^1.2 = 3.32 and b = 0.7 (the gradient itself, not e^0.7), so y = 3.32x^0.7.
Why use a logarithmic scale for very large or very small numbers?HL: Taking logs turns each power of 10 into a step of 1, so values spread over many orders of magnitude fit on one scale. Example: log₁₀ 0.00025 = −3.60.