Graphing Functions (Leaving Cert Mathematics): Model Answers

Quadratic Functions and Profit Optimization Modelling

Sample Answer

Part (a)(i) - Finding P(0) and its meaning

Working

To find $P(0)$, substitute $x = 0$ into the profit function:

$P(x) = -1.5x^2 + 10.5x - 4$

$P(0) = -1.5(0)^2 + 10.5(0) - 4$

$P(0) = 0 + 0 - 4$

$P(0) = -4$

Explanation

$P(0) = -4$ means that if the company produces zero phones in the first year, they will make a loss of €4 million. This represents the initial costs or overheads (such as setup costs, facilities, equipment) that the company incurs even before producing any phones.

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Part (a)(ii) - Completing the profit table

Working

Calculate $P(x)$ for each value of $x$ from 0 to 7:

For $x = 0$: $P(0) = -1.5(0)^2 + 10.5(0) - 4 = -4$

For $x = 1$: $P(1) = -1.5(1)^2 + 10.5(1) - 4 = -1.5 + 10.5 - 4 = 5$

For $x = 2$: $P(2) = -1.5(2)^2 + 10.5(2) - 4 = -1.5(4) + 21 - 4 = -6 + 21 - 4 = 11$

For $x = 3$: $P(3) = -1.5(3)^2 + 10.5(3) - 4 = -1.5(9) + 31.5 - 4 = -13.5 + 31.5 - 4 = 14$

For $x = 4$: $P(4) = -1.5(4)^2 + 10.5(4) - 4 = -1.5(16) + 42 - 4 = -24 + 42 - 4 = 14$

For $x = 5$: $P(5) = -1.5(5)^2 + 10.5(5) - 4 = -1.5(25) + 52.5 - 4 = -37.5 + 52.5 - 4 = 11$

For $x = 6$: $P(6) = -1.5(6)^2 + 10.5(6) - 4 = -1.5(36) + 63 - 4 = -54 + 63 - 4 = 5$

For $x = 7$: $P(7) = -1.5(7)^2 + 10.5(7) - 4 = -1.5(49) + 73.5 - 4 = -73.5 + 73.5 - 4 = -4$

Completed Table

Number of phones produced, $x$ (in tens of thousands)	0	1	2	3	4	5	6	7
Profit, P(x) (in millions of euro)	-4	5	11	14	14	11	5	-4

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Part (a)(iii) - Drawing the graph of P(x)

Description of Graph

Students should plot the following points from the completed table and draw a smooth parabola through them:

Points to plot:

• $(0, -4)$
• $(1, 5)$
• $(2, 11)$
• $(3, 14)$
• $(4, 14)$
• $(5, 11)$
• $(6, 5)$
• $(7, -4)$

Key features:

• The graph is a downward-opening parabola (negative coefficient of $x^2$)
• Maximum point occurs between $x = 3$ and $x = 4$ at approximately $(3.5, 14)$
• The curve is symmetrical about the line $x = 3.5$
• The parabola crosses the x-axis between $x = 0$ and $x = 1$, and again between $x = 6$ and $x = 7$
• All points should be accurately plotted and joined with a smooth curve (not straight line segments)

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Part (a)(iv) - Estimating range of x for profit ≥ €6 million

Working

From the graph, we need to find the range of $x$ values where $P(x) \geq 6$.

Method:

1. Draw a horizontal line at $y = 6$ on the graph
2. Find where this line intersects the parabola
3. The x-coordinates of these intersection points give the range

From the graph:

• The line $y = 6$ intersects the parabola at two points
• First intersection: approximately $x = 1.2$ (between $x = 1$ where $P(1) = 5$ and $x = 2$ where $P(2) = 11$)
• Second intersection: approximately $x = 5.8$ (between $x = 5$ where $P(5) = 11$ and $x = 6$ where $P(6) = 5$)

Students should show work on the graph by:

• Drawing a horizontal line at $P(x) = 6$
• Marking the intersection points
• Reading off the x-values

Final Answer

The company will have a profit of at least €6 million when:

$1.2 \leq x \leq 5.8$

This means the company needs to produce between 12,000 and 58,000 phones (since $x$ is in tens of thousands).

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Part (b)(i) - Finding Q'(x) and the value of x for maximum profit

Working

Step 1: Differentiate Q(x)

Given: $Q(x) = -1.5x^2 + 9.6x - 3.5$

Using the power rule: $\frac{d}{dx}(ax^n) = nax^{n-1}$

$Q'(x) = -1.5(2x) + 9.6(1) - 0$

$Q'(x) = -3x + 9.6$

Step 2: Set the derivative equal to zero

For maximum or minimum values, $Q'(x) = 0$:

$-3x + 9.6 = 0$

Step 3: Solve for x

$-3x = -9.6$

$x = \frac{-9.6}{-3}$

$x = 3.2$

Step 4: Verify this is a maximum

Since the coefficient of $x^2$ in $Q(x)$ is negative ($-1.5$), the parabola opens downward, so $x = 3.2$ gives a maximum value.

Final Answer

$Q'(x) = -3x + 9.6$

Value of x for maximum profit: $x = 3.2$ (which represents 32,000 phones)

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Part (b)(ii) - Finding the maximum value of Q(x)

Working

Substitute $x = 3.2$ into $Q(x)$:

$Q(3.2) = -1.5(3.2)^2 + 9.6(3.2) - 3.5$

Calculate $(3.2)^2$: $(3.2)^2 = 10.24$

Substitute: $Q(3.2) = -1.5(10.24) + 9.6(3.2) - 3.5$

$Q(3.2) = -15.36 + 30.72 - 3.5$

$Q(3.2) = 11.86$

Final Answer

Maximum profit in the second year: $Q(3.2) = 11.86$ million euro

This means the company will achieve a maximum profit of €11.86 million when they produce 32,000 phones in the second year.

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Part (c)(i) - Estimating R(2) and calculating R(2) + 3

Working

From the graph provided:

Looking at the graph of $R(x)$ at $x = 2$:

• Locate $x = 2$ on the horizontal axis
• Draw a vertical line up to the curve
• Read the corresponding y-value

Estimate: $R(2) \approx 8$ (the curve passes through approximately $(2, 8)$)

Calculate R(2) + 3: $R(2) + 3 = 8 + 3 = 11$

Final Answer

From the graph: $R(2) = 8$ (million euro)

Therefore: $R(2) + 3 = 11$ (million euro)

This value of 11 represents the actual profit for the third year after receiving the €3 million additional funding.

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Part (c)(ii) - Drawing the graph of y = R(x) + 3

Description of Graph

The graph of $y = R(x) + 3$ is a vertical translation of the graph of $R(x)$ upward by 3 units.

Key features to include:

1. Same shape as R(x): The new curve has identical shape to the original parabola - same width, same curvature
2. Vertical shift: Every point on $R(x)$ is moved 3 units upward:
   • If a point $(a, b)$ is on $R(x)$, then $(a, b+3)$ is on $y = R(x) + 3$
3. Key points to plot:
   • Original $R(0) \approx 0$, so new point: $(0, 3)$
   • Original $R(2) \approx 8$, so new point: $(2, 11)$ ✓ (matches our calculation)
   • Original maximum at approximately $(3, 9)$, so new maximum: $(3, 12)$
   • Original $R(6) \approx 0$, so new point: $(6, 3)$
4. Same x-axis crossings shifted: The parabola no longer crosses the x-axis since all y-values are increased by 3
5. Use the same axes and scales as provided in the question

Drawing instructions:

• Draw a smooth parabola
• Ensure it's the same shape as $R(x)$ but positioned 3 units higher
• The curve should pass through $(2, 11)$ as calculated in part (i)
• Label the curve as $y = R(x) + 3$
• Ensure smooth joining of all points (no sharp corners)

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Marking Scheme