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Functions - Modelling using functions

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Modelling involves choosing a function to represent a real-world relationship. Linear models, f(x)=mx+cf(x) = mx + c, are used when there is a constant rate of change. The gradient mm represents the rate of change (e.g., speed or cost per unit), and the yy-intercept cc represents the initial value (e.g., starting height or fixed fee).

Graph of a linear model showing a fixed intercept and a constant slope.
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Quadratic models, f(x)=ax2+bx+cf(x) = ax^2 + bx + c, represent situations involving acceleration, projectile motion, or area optimization. The vertex (−b2a,f(−b2a))(\frac{-b}{2a}, f(\frac{-b}{2a})) identifies the maximum or minimum point of the model, which is critical for optimization problems.

A downward opening parabola representing a projectile reaching a maximum height.
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Exponential models, f(x)=a⋅bxf(x) = a \cdot b^x, represent rapid growth (b>1b > 1) or decay (0<b<10 < b < 1), such as population growth, radioactive decay, or compound interest. The horizontal asymptote (usually y=0y=0 if not shifted) indicates a limit that the value approaches over time.

An exponential growth curve starting near the origin and increasing steeply.
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Sinusoidal (Periodic) models, f(t)=asin⁡(b(t−c))+df(t) = a \sin(b(t-c)) + d, are used for phenomena that repeat over a set interval, such as tides, sound waves, or Ferris wheels. Parameters define the amplitude (aa), period (360∘b\frac{360^{\circ}}{b} or 2πb\frac{2\pi}{b}), phase shift (cc), and vertical shift or principal axis (dd).

A sine wave oscillating around a horizontal principal axis.

📐Formulae

f(x)=mx+cf(x) = mx + c

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

x=−b2ax = -\frac{b}{2a}

f(x)=a⋅bxf(x) = a \cdot b^x

y=kxny = \frac{k}{x^n}

f(t)=asin⁡(b(t−c))+df(t) = a \sin(b(t - c)) + d

💡Examples

Problem 1:

A plumber charges a fixed call-out fee of 5050 plus an hourly rate of 3030. Write a linear model for the total cost CC in terms of hours hh worked, and calculate the cost for 4.54.5 hours.

Solution:

C(h)=30h+50C(h) = 30h + 50 For h=4.5h = 4.5: C(4.5)=30(4.5)+50C(4.5) = 30(4.5) + 50 C(4.5)=135+50=185C(4.5) = 135 + 50 = 185

Explanation:

The fixed fee represents the y-intercept (c=50c = 50) and the hourly rate represents the gradient (m=30m = 30). The total cost is 185185 units of currency.

Problem 2:

The height hh (in meters) of a ball thrown into the air is modelled by h(t)=−5t2+20t+2h(t) = -5t^2 + 20t + 2, where tt is the time in seconds. Find the maximum height reached by the ball.

Solution:

The maximum height occurs at the vertex. The time tt is: t=−b2a=−202(−5)=−20−10=2 secondst = -\frac{b}{2a} = -\frac{20}{2(-5)} = \frac{-20}{-10} = 2 \text{ seconds} Substitute t=2t = 2 into the function: h(2)=−5(2)2+20(2)+2h(2) = -5(2)^2 + 20(2) + 2 h(2)=−5(4)+40+2h(2) = -5(4) + 40 + 2 h(2)=−20+40+2=22 mh(2) = -20 + 40 + 2 = 22 \text{ m}

Explanation:

The maximum of a downward-opening parabola is found at its vertex. The tt-coordinate of the vertex gives the time, and the hh-coordinate gives the maximum height.

Problem 3:

The population of a city is 500,000500,000 and is growing at a rate of 3%3\% per year. Write an exponential model for the population PP after tt years.

Solution:

The initial value a=500000a = 500000. The growth factor b=1+r=1+0.03=1.03b = 1 + r = 1 + 0.03 = 1.03. P(t)=500000⋅(1.03)tP(t) = 500000 \cdot (1.03)^t

Explanation:

Exponential growth models use the formula P=a(1+r)tP = a(1+r)^t where rr is the decimal growth rate.

Problem 4:

The water level HH (in meters) in a harbor varies periodically with time tt (in hours after midnight). The model is given by H(t)=4cos⁡(30(t−2))+6H(t) = 4 \cos(30(t - 2)) + 6. Find the water level at 6:006:00 AM and state the maximum depth of the water.

Graph of the harbor water level showing a peak at 10m and a value of 4m at t=6.

Solution:

  1. To find the level at 6:006:00 AM, substitute t=6t = 6 into the equation: H(6)=4cos⁡(30(6−2))+6H(6) = 4 \cos(30(6 - 2)) + 6 H(6)=4cos⁡(30×4)+6H(6) = 4 \cos(30 \times 4) + 6 H(6)=4cos⁡(120∘)+6H(6) = 4 \cos(120^{\circ}) + 6 Since cos⁡(120∘)=−0.5\cos(120^{\circ}) = -0.5: H(6)=4(−0.5)+6=−2+6=4 mH(6) = 4(-0.5) + 6 = -2 + 6 = 4\text{ m}
  2. The maximum depth is found at the peak of the cosine wave, which is the vertical shift dd plus the amplitude aa: Max=d+a=6+4=10 m\text{Max} = d + a = 6 + 4 = 10\text{ m}

Explanation:

This is a periodic model. The cosine function oscillates between −1-1 and 11. Multiplied by the amplitude (44), it oscillates between −4-4 and 44. Adding the vertical shift (66) moves the center of the oscillation to 66, resulting in a range of [2,10][2, 10].

Problem 5:

A rectangle is inscribed inside a right-angled triangle with a base of 1010 cm and a height of 2020 cm. If the width of the rectangle is xx, show that the area AA is given by A(x)=20x−2x2A(x) = 20x - 2x^2. Find the value of xx that maximizes the area.

A right triangle containing an inscribed rectangle with width x.

Solution:

  1. Using similar triangles, the height hh of the rectangle relates to xx as: 20−hx=2010\frac{20 - h}{x} = \frac{20}{10} 20−h=2x  ⟹  h=20−2x20 - h = 2x \implies h = 20 - 2x
  2. The area A=x⋅hA = x \cdot h: A=x(20−2x)=20x−2x2A = x(20 - 2x) = 20x - 2x^2
  3. This is a quadratic with a=−2a = -2 and b=20b = 20. The maximum occurs at the vertex: x=−b2a=−202(−2)=204=5 cmx = -\frac{b}{2a} = -\frac{20}{2(-2)} = \frac{20}{4} = 5\text{ cm}

Explanation:

By expressing the dimensions of the rectangle in terms of a single variable xx using geometric properties (similar triangles), we create a quadratic function for the area. Finding the vertex of this parabola identifies the optimal width for the largest possible area.