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Practice SAC 1

Functions, Relations and Graphs

Area of Study:Functions, Relations and Graphs Task format:Test Working time:50 minutes Reading time:5 minutes Total marks:50 marks CAS calculator:Permitted

Disclaimer: This is a practice paper. Your school's SAC may differ in length, marks, and time allocation. The format and difficulty are calibrated to be representative of a typical Unit 1 SAC.

Context: Eastside Community Sports Complex

A new sports complex is being planned in the Eastside region. Architects and engineers use mathematical models to design facilities, plan budgets, and forecast membership growth. Functions are used to model everything from the depth profile of the Olympic pool to membership revenue projections and the shape of access ramps.

Part A: Membership Growth Model

8 marks

The sports complex forecasts membership growth over its first five years of operation. The number of members (in hundreds) at the end of year $t$ is given by: \[N(t) = \begin{cases} 5t & \text{if } 0 \leq t \leq 2 \\ 10 + 3(t - 2) & \text{if } 2 < t \leq 5 \end{cases}\]

Question 12 marks

Is $N(t)$ a function? Justify your answer using the definition of a function.

Yes, $N$ is a function. For each value of $t$ in the domain $[0, 5]$, there is exactly one output value $N(t)$. (1 mark: correct answer. 1 mark: valid justification using definition.)

How did you go? (out of 2)

Question 26 marks

a. Find the domain and range of $N(t)$. [2 marks]

b. Evaluate $N(2)$ and $N(4)$, and interpret the results in context. [2 marks]

c. Sketch the graph of $N(t)$ for $0 \leq t \leq 5$. Clearly show the point where the two pieces join and label the endpoints. [2 marks]

Graph space

a. [2 marks]

Domain: $[0, 5]$ (1 mark). Range: $[0, 19]$ since $N(0) = 0$ and $N(5) = 10 + 3(3) = 19$ (1 mark).

b. [2 marks]

$N(2) = 5 \times 2 = 10$ (using first piece). $N(4) = 10 + 3(4-2) = 16$ (using second piece).

$N(2) = 10$ means 1000 members after 2 years. $N(4) = 16$ means 1600 members after 4 years. (1 mark each for correct evaluation and interpretation.)

c. [2 marks]

First piece: line from $(0, 0)$ to $(2, 10)$ with gradient 5. Second piece: line from $(2, 10)$ to $(5, 19)$ with gradient 3. The pieces join continuously at $(2, 10)$.

1 mark: both line segments with correct gradients and endpoints. 1 mark: join point $(2, 10)$ clearly shown and endpoints labelled.

How did you go? (out of 6)

Part B: Access Ramp Design

4 marks

The wheelchair access ramp rises 1 metre vertically and extends 16 metres horizontally.

Question 34 marks

a. Find the gradient of the ramp. [1 mark]

b. If the ramp equation is $y = mx + c$ and the ramp starts at the origin, write the equation of the ramp. [1 mark]

c. If there is a parallel safety rail 0.2 metres above the ramp, find its equation. [1 mark]

d. Find the gradient of a line perpendicular to the ramp. [1 mark]

a. [1 mark]

$m = \frac{1}{16} = 0.0625$

b. [1 mark]

$y = \frac{1}{16}x$

c. [1 mark]

$y = \frac{1}{16}x + 0.2$ (parallel lines have same gradient)

d. [1 mark]

$m_{\perp} = -16$ (negative reciprocal)

How did you go? (out of 4)

Part C: Pool Design

12 marks

The cross-section of the 50-metre Olympic pool has a parabolic depth profile modelled by $d(x) = -\frac{1}{25}x^2 + 2x$, where $x$ is horizontal distance from one end (in metres) and $d$ is the depth of the pool (in metres). The pool extends from $x = 0$ to $x = 50$.

Question 43 marks

a. Express $d(x)$ in turning point form by completing the square. [1 mark]

b. Find the coordinates of the deepest point of the pool. [1 mark]

c. State the maximum depth of the pool. [1 mark]

a. [1 mark]

Factor out $-\frac{1}{25}$: $d(x) = -\frac{1}{25}(x^2 - 50x)$. Complete the square: $= -\frac{1}{25}(x^2 - 50x + 625 - 625) = -\frac{1}{25}((x - 25)^2 - 625) = -\frac{1}{25}(x - 25)^2 + 25$

$d(x) = -\frac{1}{25}(x - 25)^2 + 25$

b. [1 mark]

The turning point form $d(x) = -\frac{1}{25}(x - 25)^2 + 25$ has turning point at $(25, 25)$. Since the coefficient of $(x-25)^2$ is negative, this is a maximum.

Deepest point: $(25, 25)$

c. [1 mark]

The maximum depth is 25 metres (at $x = 25$, the centre of the pool).

How did you go? (out of 3)

Question 53 marks

Solve $-\frac{1}{25}x^2 + 2x = 5$ to determine the horizontal distances at which the pool is 5 metres deep. Give your answers correct to two decimal places.

\[-\frac{1}{25}x^2 + 2x = 5\] \[-\frac{1}{25}x^2 + 2x - 5 = 0\] Multiply both sides by $-25$: \[x^2 - 50x + 125 = 0\]

Using quadratic formula: $x = \frac{50 \pm \sqrt{2500 - 500}}{2} = \frac{50 \pm \sqrt{2000}}{2} = \frac{50 \pm 44.72}{2}$

$x \approx 2.64$ metres and $x \approx 47.36$ metres

1 mark: correct equation setup. 1 mark: correct application of formula. 1 mark: correct answers to 2 d.p.

How did you go? (out of 3)

Question 66 marks

a. An engineer proposes deepening the pool so its maximum depth reaches 30 metres. Use the discriminant to determine whether the equation $d(x) = 30$ has any real solutions. [1 mark]

b. Interpret this result in the context of the pool design. [1 mark]

c. Sketch the graph of $d(x) = -\frac{1}{25}x^2 + 2x$ for $0 \leq x \leq 50$. Label the turning point and intercepts. [2 marks]

Graph space

d. For what horizontal distances is the pool deeper than 20 metres? Express your answer as an inequality. [2 marks]

a. [1 mark]

$-\frac{1}{25}x^2 + 2x = 30$ gives $-\frac{1}{25}x^2 + 2x - 30 = 0$, or $x^2 - 50x + 750 = 0$.

$\Delta = (-50)^2 - 4(1)(750) = 2500 - 3000 = -500$

$\Delta = -500 < 0$, so there are no real solutions.

b. [1 mark]

The pool's depth never reaches 30 metres with the current profile. The maximum depth is 25 metres (at the centre), so a depth of 30 metres is not achievable without changing the model.

c. [2 marks]

x-intercepts: $d(x) = 0 \Rightarrow -\frac{1}{25}x(x - 50) = 0$, so $x = 0$ and $x = 50$. Turning point: $(25, 25)$. Inverted parabola (opens downward).

1 mark: correct shape (inverted parabola) with turning point $(25, 25)$ labelled. 1 mark: x-intercepts $(0, 0)$ and $(50, 0)$ labelled.

d. [2 marks]

Solve $d(x) > 20$: $-\frac{1}{25}x^2 + 2x > 20$. Rearrange: $-\frac{1}{25}x^2 + 2x - 20 > 0$. Multiply by $-25$ (reverse inequality): $x^2 - 50x + 500 < 0$.

Roots: $x = \frac{50 \pm \sqrt{2500 - 2000}}{2} = \frac{50 \pm \sqrt{500}}{2} \approx \frac{50 \pm 22.36}{2}$. So $x \approx 13.82$ or $x \approx 36.18$.

$13.82 < x < 36.18$ (the pool is deeper than 20 metres between approximately 13.82 m and 36.18 m from the end).

1 mark: correct setup and solving the quadratic. 1 mark: correct inequality answer.

How did you go? (out of 6)

Part D: Revenue Model

9 marks

The weekly revenue (in thousands of dollars) depends on the number of fitness classes offered. A model is: \[R(c) = c^3 - 12c^2 + 36c\] where $c$ is the number of classes per week and $0 \leq c \leq 10$.

Question 73 marks

a. Factorise $R(c) = c^3 - 12c^2 + 36c$. [1 mark]

b. Find the x-intercepts of the function. [1 mark]

c. Interpret the x-intercepts in the context of revenue. [1 mark]

a. [1 mark]

$R(c) = c(c^2 - 12c + 36) = c(c - 6)^2$

b. [1 mark]

$c = 0$ (simple root) and $c = 6$ (double root)

c. [1 mark]

At $c = 0$, zero classes means zero revenue. At $c = 6$, the curve touches but does not cross the axis (repeated root), indicating a local minimum in revenue where revenue drops back to zero.

How did you go? (out of 3)

Question 86 marks

a. Use your CAS calculator to find the coordinates of the two turning points of $R(c)$. [2 marks]

b. Classify each turning point as a local maximum or local minimum, and state the maximum weekly revenue. [2 marks]

c. Sketch the graph of $R(c)$ for $0 \leq c \leq 10$, labelling all intercepts and turning points. [1 mark]

Graph space

d. State the practical domain and range of $R(c)$, given that the complex can offer a maximum of 10 classes per week and revenue cannot be negative. [1 mark]

a. [2 marks]

Turning points: $(2, 32)$ and $(6, 0)$. (1 mark for each correct turning point.)

b. [2 marks]

$(2, 32)$ is a local maximum. $(6, 0)$ is a local minimum. Maximum weekly revenue is \$32{,}000 (at 2 classes per week). (1 mark for classification, 1 mark for stating max revenue in context.)

c. [1 mark]

x-intercepts at $c = 0$ (simple root - crosses axis) and $c = 6$ (double root - touches axis). Local max at $(2, 32)$, local min at $(6, 0)$. At $c = 10$: $R(10) = 10(4)^2 = 160$. Cubic shape, positive leading coefficient.

1 mark: correct cubic shape with intercepts and turning points labelled.

d. [1 mark]

Domain: $[0, 10]$. Range: $[0, 160]$. ($R(10) = 10 \times 4^2 = 160$ is the maximum value on this domain.)

How did you go? (out of 6)

Part E: Energy and Cooling

9 marks

The cooling power of the HVAC system is modelled by $P(T) = \frac{k}{T}$, where $T$ is ambient temperature (in degrees Celsius above 15) and $P$ is power consumption (in kilowatts).

Question 94 marks

a. When $T = 5$, power consumption is 12 kW. Find the constant $k$. [1 mark]

b. Write the complete rule for $P(T)$. [1 mark]

c. Find $P(10)$ and interpret in context. [1 mark]

d. Describe the end behaviour of $P(T)$ as $T \to \infty$. [1 mark]

a. [1 mark]

$12 = \frac{k}{5} \Rightarrow k = 60$

b. [1 mark]

$P(T) = \frac{60}{T}$

c. [1 mark]

$P(10) = 6$ kW. When the ambient temperature is 25C (10 above 15C), the cooling system uses 6 kW.

d. [1 mark]

As $T \to \infty$, $P(T) \to 0^+$. Power consumption approaches zero (but remains positive).

How did you go? (out of 4)

Question 105 marks

The volume of water (in cubic metres) as the pool fills is modelled by $V(t) = 2t^{3/2}$, where $t$ is time in hours after the pump is turned on.

a. Evaluate $V(1)$ and $V(4)$. [1 mark]

b. How long does it take to fill 54 cubic metres? Solve $V(t) = 54$. [2 marks]

c. State the domain and range of $V(t)$ if the pool holds a maximum of 128 cubic metres. [1 mark]

d. Explain why the rate at which the pool fills increases over time. Refer to the index of the power function in your answer. [1 mark]

a. [1 mark]

$V(1) = 2 \times 1^{3/2} = 2$. $V(4) = 2 \times 4^{3/2} = 2 \times 8 = 16$.

$V(1) = 2$ m$^3$ and $V(4) = 16$ m$^3$.

b. [2 marks]

$2t^{3/2} = 54 \Rightarrow t^{3/2} = 27$. Raise both sides to the power $\frac{2}{3}$: $t = 27^{2/3} = (27^{1/3})^2 = 3^2 = 9$.

It takes 9 hours to fill 54 m$^3$. (1 mark: correct setup. 1 mark: correct answer.)

c. [1 mark]

$V(t) = 128 \Rightarrow 2t^{3/2} = 128 \Rightarrow t^{3/2} = 64 \Rightarrow t = 64^{2/3} = (64^{1/3})^2 = 4^2 = 16$.

Domain: $[0, 16]$. Range: $[0, 128]$.

d. [1 mark]

The index $\frac{3}{2} > 1$, so the graph is concave up (increasing at an increasing rate). This means the pool fills faster as time goes on, unlike a linear model where the rate would be constant.

How did you go? (out of 5)

Part F: Function Transformations

8 marks

The membership forecasting model is based on a simple function $f(x) = x^2$ that is transformed to fit actual data.

Question 112 marks

Describe, in sequence, the transformations applied to $y = x^2$ to obtain $y = 2(x - 1)^2 + 3$.

1. Translate right 1 unit: $(x - 1)^2$. 2. Dilate vertically by factor 2: $2(x - 1)^2$. 3. Translate up 3 units: $2(x - 1)^2 + 3$. (1 mark for sequence, 1 mark for correct descriptions.)

How did you go? (out of 2)

Question 123 marks

a. A function $g(x) = (x + 2)^2$ is the result of translating $f(x) = x^2$. Describe the translation. [1 mark]

b. A function $h(x) = \frac{1}{2}x^2$ is obtained by dilating $f(x) = x^2$ parallel to the y-axis. State the scale factor. [1 mark]

c. Write the rule for the function obtained by reflecting $f(x) = x^2$ in the x-axis. [1 mark]

a. [1 mark]

Translation left 2 units (or by vector (-2, 0))

b. [1 mark]

Scale factor = $\frac{1}{2}$ (vertical compression by factor 2)

c. [1 mark]

$y = -x^2$

How did you go? (out of 3)

Question 133 marks

A function $y = (x - 2)^2 - 4$ is shown on a graph. Write the rule for the function after applying: (i) a translation 3 units right, (ii) followed by a dilation of factor 2 parallel to the y-axis.

Start: $y = (x - 2)^2 - 4$

After translating right 3: Replace $x$ with $x - 3$: $y = ((x-3) - 2)^2 - 4 = (x - 5)^2 - 4$

After dilating by 2 parallel to y-axis: Multiply output by 2: $y = 2[(x - 5)^2 - 4] = 2(x - 5)^2 - 8$

$y = 2(x - 5)^2 - 8$

1 mark: correct translation rule. 1 mark: correct dilation application. 1 mark: final answer.

How did you go? (out of 3)

End of Questions

Before You Start

Practice SAC 1

Study Mode

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Practice SAC 1

Part A: Membership Growth Model

Part B: Access Ramp Design

Part C: Pool Design

Part D: Revenue Model

Part E: Energy and Cooling

Part F: Function Transformations

Performance Review