Secondary 3 Elementary Mathematics Graphs Coordinate Geometry Quiz

Secondary 3 Elementary Mathematics Quiz - Answers

Graphs Coordinate Geometry

Total Marks: 50

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Section A: Coordinate Geometry Foundations

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1. Find the distance between $A(3, -2)$ and $B(7, 4)$. [2 marks]

Method: Use the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Step-by-step:

• $d = \sqrt{(7-3)^2 + (4-(-2))^2}$ [1 mark for correct substitution]
• $d = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$ [1 mark]
• $= \sqrt{4 \times 13} = 2\sqrt{13}$ units

Answer: $2\sqrt{13}$ units (or $\sqrt{52}$ units)

Teaching note: The distance formula comes from Pythagoras' theorem. The horizontal distance is $\Delta x$ and vertical distance is $\Delta y$, so the direct distance is the hypotenuse. Always simplify surds by extracting square factors.

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2. Find the midpoint of $P(-5, 8)$ and $Q(3, -4)$. [2 marks]

Method: Use the midpoint formula: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

Step-by-step:

• Midpoint $= \left(\frac{-5+3}{2}, \frac{8+(-4)}{2}\right)$ [1 mark]
• $= \left(\frac{-2}{2}, \frac{4}{2}\right)$ [0.5 mark]
• $= (-1, 2)$ [0.5 mark]

Answer: $(-1, 2)$

Teaching note: The midpoint is simply the average of the $x$-coordinates and the average of the $y$-coordinates. This represents the "centre point" of the line segment.

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3. The gradient of $C(2, k)$ to $D(6, 10)$ is $2$. Find $k$. [2 marks]

Method: Use gradient formula: $m = \frac{y_2-y_1}{x_2-x_1}$

Step-by-step:

• $2 = \frac{10-k}{6-2} = \frac{10-k}{4}$ [1 mark for setup]
• $8 = 10 - k$ [0.5 mark]
• $k = 10 - 8 = 2$ [0.5 mark]

Answer: $k = 2$

Teaching note: Gradient measures "rise over run" — how much $y$ changes for each unit change in $x$. A gradient of 2 means $y$ increases by 2 when $x$ increases by 1.

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4. Find the gradient of $3x - 4y + 12 = 0$. [2 marks]

Method: Rearrange to $y = mx + c$ form.

Step-by-step:

• $3x - 4y + 12 = 0$
• $4y = 3x + 12$ [0.5 mark]
• $y = \frac{3}{4}x + 3$ [1 mark for correct rearrangement]
• Gradient $m = \frac{3}{4}$ [0.5 mark]

Answer: $\frac{3}{4}$ (or $0.75$)

Teaching note: The coefficient of $x$ in $y = mx + c$ is always the gradient. When rearranging, be careful with signs — dividing by negative requires flipping all signs.

Common mistake: Forgetting to change signs when moving terms across the equals sign, or dividing only part of the equation by the coefficient of $y$.

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5. Equation of line through $(5, -1)$ with gradient $-\frac{2}{3}$. [2 marks]

Method: Use point-gradient form $y - y_1 = m(x - x_1)$, then rearrange.

Step-by-step:

• $y - (-1) = -\frac{2}{3}(x - 5)$ [1 mark for correct substitution]
• $y + 1 = -\frac{2}{3}x + \frac{10}{3}$
• $y = -\frac{2}{3}x + \frac{10}{3} - 1 = -\frac{2}{3}x + \frac{10}{3} - \frac{3}{3}$ [0.5 mark]
• $y = -\frac{2}{3}x + \frac{7}{3}$ [0.5 mark]

Answer: $y = -\frac{2}{3}x + \frac{7}{3}$ (or $y = -\frac{2}{3}x + 2\frac{1}{3}$)

Teaching note: The point-gradient form is most efficient when you know one point and the gradient. The $y$-intercept $c$ can be found by substitution: $-1 = -\frac{2}{3}(5) + c$ gives $c = \frac{7}{3}$.

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6. Equation through $(-2, 4)$ and $(4, -2)$. [2 marks]

Method: Find gradient first, then use point-gradient form.

Step-by-step:

• Gradient $m = \frac{-2-4}{4-(-2)} = \frac{-6}{6} = -1$ [0.5 mark]
• Using point $(4, -2)$: $y - (-2) = -1(x - 4)$ [0.5 mark]
• $y + 2 = -x + 4$
• $y = -x + 2$ [0.5 mark for equation]
• $x + y - 2 = 0$ [0.5 mark for required form]

Answer: $x + y - 2 = 0$ (or equivalent integer form)

Teaching note: Always check by substituting both original points into your final equation. Both should satisfy it. For $ax + by + c = 0$ with integer coefficients, eliminate fractions and ensure $a > 0$ conventionally.

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7. State whether $L_1: y = 2x + 5$ and $L_2: y = 2x - 3$ are parallel, perpendicular, or neither. [2 marks]

Answer: Parallel [1 mark]

Reason: Both lines have the same gradient $m = 2$ [1 mark]

Teaching note: Parallel lines have equal gradients ($m_1 = m_2$). Perpendicular lines have $m_1 \times m_2 = -1$ (negative reciprocals). Here $2 \times 2 = 4 \neq -1$, so not perpendicular. The different $y$-intercepts ($5 \neq -3$) confirm they are distinct parallel lines, not the same line.

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8. Equation of line perpendicular to $y = \frac{1}{2}x + 3$ through $(4, -1)$. [2 marks]

Method: Negative reciprocal gradient, then use point-gradient form.

Step-by-step:

• Gradient of given line: $\frac{1}{2}$
• Perpendicular gradient: $-2$ (negative reciprocal: $-\frac{1}{\frac{1}{2}} = -2$) [0.5 mark]
• $y - (-1) = -2(x - 4)$ [0.5 mark]
• $y + 1 = -2x + 8$
• $y = -2x + 7$ [1 mark]

Answer: $y = -2x + 7$

Teaching note: The negative reciprocal rule: if $m_1 \times m_2 = -1$, the lines are perpendicular. For $m_1 = \frac{1}{2}$, we need $m_2 = -2$ since $\frac{1}{2} \times (-2) = -1$. A quick check: flip the fraction and change the sign.

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Section B: Applications and Problem Solving

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9. Rectangle $ABCD$ with $A(1, 2)$, $B(5, 2)$, $C(5, 5)$.

(a) Find coordinates of $D$. [1 mark]

Method: In a rectangle, opposite sides are equal and parallel. $AB$ is horizontal, $BC$ is vertical.

Step-by-step:

• $AB$ goes from $x=1$ to $x=5$ at $y=2$; length 4, direction right
• $BC$ goes from $y=2$ to $y=5$ at $x=5$; length 3, direction up
• To close the rectangle from $A$ to $D$: go up 3 units: $D = (1, 5)$ [1 mark]

Answer (a): $D(1, 5)$

(b) Area of rectangle $ABCD$. [1 mark]

Method: Area = length $\times$ width

Step-by-step:

• Length $AB = 5 - 1 = 4$
• Width $BC = 5 - 2 = 3$
• Area $= 4 \times 3 = 12$ square units [1 mark]

Answer (b): 12 square units

(c) Length of diagonal $AC$. [2 marks]

Method: Distance formula.

Step-by-step:

• $AC = \sqrt{(5-1)^2 + (5-2)^2}$ [1 mark]
• $= \sqrt{16 + 9} = \sqrt{25} = 5$ [1 mark]

Answer (c): 5 units

Teaching note: Note this is a 3-4-5 right triangle, a Pythagorean triple. The diagonal of a rectangle can always be found using Pythagoras on its side lengths.

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10. Quadrilateral $P(-3, 1)$, $Q(2, 5)$, $R(6, 2)$, $S(1, -2)$.

(a) Show $PQ$ is parallel to $SR$. [2 marks]

Method: Show gradients are equal.

Step-by-step:

• Gradient of $PQ = \frac{5-1}{2-(-3)} = \frac{4}{5}$ [1 mark]
• Gradient of $SR = \frac{-2-2}{1-6} = \frac{-4}{-5} = \frac{4}{5}$ [1 mark]

Since gradients are equal, $PQ \parallel SR$.

(b) Show $PQ = SR$. [2 marks]

Method: Calculate lengths using distance formula.

Step-by-step:

• $PQ = \sqrt{(2-(-3))^2 + (5-1)^2} = \sqrt{25 + 16} = \sqrt{41}$ [1 mark]
• $SR = \sqrt{(6-1)^2 + (2-(-2))^2} = \sqrt{25 + 16} = \sqrt{41}$ [1 mark]

Therefore $PQ = SR$.

(c) Type of quadrilateral. [1 mark]

Answer: Parallelogram [0.5 mark]

Reason: One pair of opposite sides is both equal and parallel [0.5 mark]

Teaching note: A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. To prove it's specifically a rhombus, rectangle, or square requires additional conditions (all sides equal, right angles, etc.).

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11. Line $L$: $2x + 5y = 10$.

(a) Find intercepts. [2 marks]

Step-by-step:

• $x$-intercept: set $y = 0$: $2x = 10$, so $x = 5$. Point: $(5, 0)$ [1 mark]
• $y$-intercept: set $x = 0$: $5y = 10$, so $y = 2$. Point: $(0, 2)$ [1 mark]

Answer (a): $x$-intercept: $(5, 0)$; $y$-intercept: $(0, 2)$

(b) Sketch on axes. [2 marks]

Marking: [2 marks for correctly drawn line through $(5, 0)$ and $(0, 2)$ with both intercepts labelled]

Teaching note: The intercept form $\frac{x}{a} + \frac{y}{b} = 1$ is useful for quick sketching. Here: $\frac{x}{5} + \frac{y}{2} = 1$. Always label intercepts clearly on sketches.

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12. Circle with centre $C(1, 2)$, point $A(4, -3)$ on circle.

(a) Find radius. [2 marks]

Method: Radius = distance from centre to point on circle.

Step-by-step:

• $r = \sqrt{(4-1)^2 + (-3-2)^2}$ [1 mark]
• $= \sqrt{9 + 25} = \sqrt{34}$ [1 mark]

Answer (a): $\sqrt{34}$ units

(b) Equation of circle. [1 mark]

Method: $(x-a)^2 + (y-b)^2 = r^2$ where $(a,b)$ is centre.

Step-by-step:

• $(x-1)^2 + (y-2)^2 = (\sqrt{34})^2 = 34$ [1 mark]

Answer (b): $(x-1)^2 + (y-2)^2 = 34$

Teaching note: The standard form of a circle equation directly encodes the centre (with sign change) and the square of the radius. Remember: it's $r^2$ on the right side, not $r$.

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13. Quadrilateral $ABCD$ with $A(0, 3)$, $B(4, 1)$, $C(2, -3)$, $D(-2, -1)$.

(a) Gradient of $AB$. [1 mark]

• $m_{AB} = \frac{1-3}{4-0} = \frac{-2}{4} = -\frac{1}{2}$

Answer (a): $-\frac{1}{2}$

(b) Gradient of $CD$. [1 mark]

• $m_{CD} = \frac{-1-(-3)}{-2-2} = \frac{2}{-4} = -\frac{1}{2}$

Answer (b): $-\frac{1}{2}$

(c) Explain why $ABCD$ is a parallelogram. [2 marks]

Answer:

• Gradient of $AB$ = gradient of $CD = -\frac{1}{2}$, so $AB \parallel CD$ [1 mark]
• Need to also show $AD \parallel BC$:
  • $m_{AD} = \frac{-1-3}{-2-0} = \frac{-4}{-2} = 2$
  • $m_{BC} = \frac{-3-1}{2-4} = \frac{-4}{-2} = 2$ [0.5 mark]
• Both pairs of opposite sides parallel, so $ABCD$ is a parallelogram [0.5 mark]

(d) Show $ABCD$ is not a rhombus. [2 marks]

Method: Show adjacent sides are not equal, or show diagonals are not perpendicular.

Step-by-step:

• $AB = \sqrt{(4-0)^2 + (1-3)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$ [0.5 mark]
• $BC = \sqrt{(2-4)^2 + (-3-1)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ [0.5 mark]
• Actually equal... Check $AC$ and $BD$ (diagonals):
  • $AC = \sqrt{(2-0)^2 + (-3-3)^2} = \sqrt{4 + 36} = \sqrt{40}$
  • $BD = \sqrt{(-2-4)^2 + (-1-1)^2} = \sqrt{36 + 4} = \sqrt{40}$ [0.5 mark]
• Or check: $m_{AB} \times m_{BC} = -\frac{1}{2} \times 2 = -1$, so adjacent sides are perpendicular!
• This means $ABCD$ is actually a rectangle. Check: is it a square?
  • $AB = BC$? No wait, $AB = \sqrt{20}$, need to recheck $AD$:
  • $AD = \sqrt{(-2-0)^2 + (-1-3)^2} = \sqrt{4 + 16} = \sqrt{20}$
  • All sides equal! It's a rhombus. And with perpendicular adjacent sides, it's a square.

Correction for answer key: Let me recalculate properly.

Actually: $AB = \sqrt{16+4} = \sqrt{20}$, $BC = \sqrt{4+16} = \sqrt{20}$, $CD = \sqrt{4+4} = \sqrt{8}$? No...

$CD = \sqrt{(-2-2)^2 + (-1-(-3))^2} = \sqrt{16 + 4} = \sqrt{20}$

$DA = \sqrt{(0-(-2))^2 + (3-(-1))^2} = \sqrt{4 + 16} = \sqrt{20}$

All four sides equal = rhombus. And adjacent sides perpendicular = square.

Revised marking for (d): Since this is a square (special rhombus), let's use a different approach - check diagonals not perpendicular for general rhombus, or...

Actually, let me verify: $m_{AC} = \frac{-3-3}{2-0} = -3$, $m_{BD} = \frac{-1-1}{-2-4} = \frac{-2}{-6} = \frac{1}{3}$

$m_{AC} \times m_{BD} = -3 \times \frac{1}{3} = -1$, so diagonals are perpendicular.

This IS a rhombus (actually a square). The question as designed needs adjustment in future versions.

Modified answer for (d) to match intended difficulty:

To show not a rhombus, we need different coordinates. Given the question as stated, students should find:

Alternative approach for (d): Since all sides equal $\sqrt{20}$, it IS a rhombus. The question contains an error. A correct version would use $D(-1, -2)$ giving $CD \neq AB$.

** grading note:** Award full marks to any student who correctly shows all sides equal and identifies it as a rhombus/square, or who correctly identifies it's not a rhombus if they made an arithmetic error that leads to unequal sides.

For this answer key, assume the question intended non-rhombus:

Expected student method for (d):

• Find $AD = \sqrt{(-2-0)^2 + (-1-3)^2} = \sqrt{4+16} = \sqrt{20}$ [0.5 mark]
• Since $AB = AD$ and all sides appear equal, this would be a rhombus [1 mark for identifying error or correct conclusion]
• To not be a rhombus, need $AB \neq BC$: but here $AB = BC = \sqrt{20}$

Resolution: Accept "ABCD is a rhombus (in fact a square)" as correct observation. The question as written produces a square.

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14. $L_1$ through $(0, 4)$ and $(3, 0)$; $L_2$ perpendicular to $L_1$ through $(6, 5)$.

(a) Equation of $L_1$ in $ax + by + c = 0$. [2 marks]

Step-by-step:

• Gradient of $L_1 = \frac{0-4}{3-0} = -\frac{4}{3}$ [0.5 mark]
• Using $(0, 4)$: $y = -\frac{4}{3}x + 4$ [0.5 mark]
• Multiply by 3: $3y = -4x + 12$ [0.5 mark]
• $4x + 3y - 12 = 0$ [0.5 mark]

Answer (a): $4x + 3y - 12 = 0$

(b) Equation of $L_2$ in $y = mx + c$. [2 marks]

Step-by-step:

• Perpendicular gradient: $m_2 = \frac{3}{4}$ (negative reciprocal of $-\frac{4}{3}$) [0.5 mark]
• $y - 5 = \frac{3}{4}(x - 6)$ [0.5 mark]
• $y = \frac{3}{4}x - \frac{18}{4} + 5 = \frac{3}{4}x - \frac{9}{2} + \frac{10}{2}$ [0.5 mark]
• $y = \frac{3}{4}x + \frac{1}{2}$ [0.5 mark]

Answer (b): $y = \frac{3}{4}x + \frac{1}{2}$

(c) Intersection of $L_1$ and $L_2$. [2 marks]

Method: Solve simultaneously.

Step-by-step:

• From $L_2$: substitute into $L_1$ (rearranged as $y = -\frac{4}{3}x + 4$): [0.5 mark for method]
• $\frac{3}{4}x + \frac{1}{2} = -\frac{4}{3}x + 4$
• Multiply by 12: $9x + 6 = -16x + 48$ [0.5 mark]
• $25x = 42$
• $x = \frac{42}{25} = 1.68$ [0.5 mark]
• $y = \frac{3}{4} \times \frac{42}{25} + \frac{1}{2} = \frac{126}{100} + \frac{50}{100} = \frac{176}{100} = 1.76$ or use exact: $y = \frac{3}{4} \times \frac{42}{25} + \frac{1}{2} = \frac{63}{50} + \frac{25}{50} = \frac{88}{50} = \frac{44}{25}$

Check in $L_1$: $y = -\frac{4}{3} \times \frac{42}{25} + 4 = -\frac{168}{75} + \frac{300}{75} = \frac{132}{75} = \frac{44}{25}$ ✓

Answer (c): $\left(\frac{42}{25}, \frac{44}{25}\right)$ or $(1.68, 1.76)$

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15. Triangle $A(-1, 2)$, $B(5, 8)$, $C(7, 0)$.

(a) Equation of median from $A$ to midpoint of $BC$. [3 marks]

Method: Find midpoint of $BC$, then find equation through $A$ and this midpoint.

Step-by-step:

• Midpoint of $BC = \left(\frac{5+7}{2}, \frac{8+0}{2}\right) = (6, 4)$ [1 mark]
• Gradient of median $= \frac{4-2}{6-(-1)} = \frac{2}{7}$ [1 mark]
• Equation: $y - 2 = \frac{2}{7}(x - (-1))$
• $y - 2 = \frac{2}{7}(x + 1)$
• $y = \frac{2}{7}x + \frac{2}{7} + 2 = \frac{2}{7}x + \frac{16}{7}$ [1 mark]

Answer (a): $y = \frac{2}{7}x + \frac{16}{7}$ (or $7y = 2x + 16$)

(b) Coordinates of centroid. [2 marks]

Method: The centroid is at $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$, or find intersection of two medians.

Step-by-step:

• Using formula: $\left(\frac{-1+5+7}{3}, \frac{2+8+0}{3}\right) = \left(\frac{11}{3}, \frac{10}{3}\right)$ [2 marks]

Or verify with another median:

• Midpoint of $AC = (3, 1)$, gradient from $B = \frac{1-8}{3-5} = \frac{-7}{-2} = \frac{7}{2}$
• Equation: $y - 8 = \frac{7}{2}(x - 5)$
• Intersection with $y = \frac{2}{7}x + \frac{16}{7}$: solve to get same point.

Answer (b): $\left(\frac{11}{3}, \frac{10}{3}\right)$ or $(3\frac{2}{3}, 3\frac{1}{3})$

Teaching note: The centroid divides each median in ratio $2:1$ from the vertex. It's the "centre of mass" of the triangle. The formula averages all three vertices.

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Section C: Graphs of Functions

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16. Graph of $y = (x-2)^2 - 3$.

(a) Coordinates of vertex $V$. [1 mark]

Answer (a): $(2, -3)$

Teaching note: For $y = (x-p)^2 + q$, the vertex is at $(p, q)$. The value $p = 2$ gives the axis of symmetry, and $q = -3$ is the minimum value (since the coefficient of the squared term is positive).

(b) $x$-intercepts. [2 marks]

Method: Set $y = 0$ and solve.

Step-by-step:

• $(x-2)^2 - 3 = 0$ [0.5 mark]
• $(x-2)^2 = 3$ [0.5 mark]
• $x - 2 = \pm\sqrt{3}$ [0.5 mark]
• $x = 2 \pm \sqrt{3}$ [0.5 mark]

Points: $(2 + \sqrt{3}, 0)$ and $(2 - \sqrt{3}, 0)$

Approximately: $(3.73, 0)$ and $(0.27, 0)$

Answer (b): $(2 + \sqrt{3}, 0)$ and $(2 - \sqrt{3}, 0)$

(c) Equation of line of symmetry. [1 mark]

Answer (c): $x = 2$

Teaching note: The line of symmetry for a parabola in vertex form always passes through the $x$-coordinate of the vertex. It's a vertical line $x = p$.

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17. Graph of $y = \frac{2}{x}$ for $x > 0$.

(a) Gradient of chord $PQ$ where $P(1, 2)$ and $Q(2, 1)$. [2 marks]

Step-by-step:

• Gradient $= \frac{1-2}{2-1} = \frac{-1}{1} = -1$ [2 marks]

Answer (a): $-1$

(b) Estimate gradient of tangent at $R$ where $x = 1.5$, using chord from $x = 1.4$ to $x = 1.6$. [2 marks]

Step-by-step:

• At $x = 1.4$: $y = \frac{2}{1.4} = \frac{20}{14} = \frac{10}{7} \approx 1.429$ [0.5 mark]
• At $x = 1.6$: $y = \frac{2}{1.6} = \frac{20}{16} = 1.25$ [0.5 mark]
• Gradient of chord $= \frac{1.25 - \frac{10}{7}}{1.6 - 1.4} = \frac{\frac{5}{4} - \frac{10}{7}}{0.2} = \frac{\frac{35-40}{28}}{0.2} = \frac{-5/28}{1/5} = -\frac{25}{28} \approx -0.893$ [1 mark]

Or numerically: $\frac{1.25 - 1.4286}{0.2} = \frac{-0.1786}{0.2} \approx -0.89$

Answer (b): Approximately $-0.89$ (accept $-\frac{25}{28}$ or approximately $-0.9$)

Teaching note: This is the fundamental idea behind differentiation from first principles — the gradient of a chord approaches the gradient of the tangent as the points get closer together. For $y = \frac{2}{x}$, the exact derivative is $-\frac{2}{x^2}$, giving $-\frac{2}{2.25} = -\frac{8}{9} \approx -0.889$ at $x = 1.5$.

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18. Sketch $y = 2^x$ and $y = 2^{-x}$ for $-3 \leq x \leq 3$.

(a) Label each curve. [2 marks]

Marking: [1 mark for correct exponential growth curve $y = 2^x$ passing through $(0,1)$, increasing; 1 mark for correct exponential decay curve $y = 2^{-x}$ passing through $(0,1)$, decreasing]

Key points for $y = 2^x$:

• $(-3, \frac{1}{8})$, $(-2, \frac{1}{4})$, $(-1, \frac{1}{2})$, $(0, 1)$, $(1, 2)$, $(2, 4)$, $(3, 8)$

Key points for $y = 2^{-x} = (\frac{1}{2})^x$:

• $(-3, 8)$, $(-2, 4)$, $(-1, 2)$, $(0, 1)$, $(1, \frac{1}{2})$, $(2, \frac{1}{4})$, $(3, \frac{1}{8})$

(b) Coordinates of intersection. [1 mark]

Answer (b): $(0, 1)$

Teaching note: Both curves pass through $(0, 1)$ since $2^0 = 1$ and $2^{-0} = 1$. The curves are reflections of each other in the $y$-axis. Exponential functions $y = a^x$ always pass through $(0, 1)$ for $a > 0$.

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19. Graph of $y = x^3 - 3x + 1$.

(a) Estimate solutions to $x^3 - 3x + 1 = 0$. [3 marks]

Method: Read $x$-intercepts from graph.

Expected values from graph description:

• Left intercept: approximately $x \approx -1.9$ or $-1.8$ [1 mark]
• Middle intercept: approximately $x \approx 0.3$ or $0.4$ [1 mark]
• Right intercept: approximately $x \approx 1.5$ or $1.6$ [1 mark]

More precise values: $x \approx -1.88$, $0.35$, $1.53$

Acceptable range:

• First root: $-2.0$ to $-1.8$
• Second root: $0.2$ to $0.5$
• Third root: $1.4$ to $1.7$

Answer (a): $x \approx -1.9$, $x \approx 0.3$, $x \approx 1.5$ (acceptable ranges apply)

(b) Estimate solutions to $x^3 - 4x + 1 = 0$ by drawing suitable line. [2 marks]

Method: Rewrite as $x^3 - 3x + 1 = x$, so draw line $y = x$ and find intersections.

Step-by-step:

• $x^3 - 4x + 1 = 0$
• $x^3 - 3x + 1 = x$ [1 mark for identifying line $y = x$]
• Draw line $y = x$ on graph
• Intersections give solutions: approximately $x \approx -1.9$, $x \approx 0.25$, $x \approx 1.7$ [1 mark for three reasonable estimates]

Teaching note: This technique of rewriting equations to use existing graphs is powerful. To solve $f(x) = g(x)$, you can either graph $y = f(x)$ and $y = g(x)$ and find intersections, or rewrite as $f(x) - g(x) = 0$ and find roots.

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20. Quadratic $y = -(x-p)^2 + q$, maximum value 5, passes through $(1, 1)$.

(a) Find $p$ and $q$. [3 marks]

Step-by-step:

• Maximum value is $q = 5$ (since the negative square term is always $\leq 0$) [1 mark]
• So $y = -(x-p)^2 + 5$ [0.5 mark]
• Passes through $(1, 1)$: $1 = -(1-p)^2 + 5$ [0.5 mark]
• $(1-p)^2 = 4$ [0.5 mark]
• $1-p = \pm 2$
• $p = 1 \mp 2$, so $p = 3$ or $p = -1$
• Since $p$ is positive, $p = 3$ [0.5 mark]

Answer (a): $p = 3$, $q = 5$

(b) Find $x$-intercepts. [2 marks]

Step-by-step:

• $y = -(x-3)^2 + 5$ [0.5 mark]
• Set $y = 0$: $(x-3)^2 = 5$ [0.5 mark]
• $x - 3 = \pm\sqrt{5}$ [0.5 mark]
• $x = 3 \pm \sqrt{5}$ [0.5 mark]

Answer (b): $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}$ (or approximately $5.24$ and $0.76$)

Teaching note: The negative sign before the squared term means the parabola opens downward, giving a maximum at the vertex. Always check that your value of $p$ satisfies all given conditions — if the question hadn't specified $p > 0$, both $p = 3$ and $p = -1$ would be mathematically valid, giving different but related parabolas.

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END OF ANSWER KEY