O Level Elementary Mathematics Graphs Coordinate Geometry Quiz

O-Level Elementary Mathematics Quiz - Graphs Coordinate Geometry (Answer Key)

Total Marks: 40
Topic: Graphs & Coordinate Geometry
Level: O-Level Elementary Mathematics

Section A: Coordinates and Basic Graphs

Q1 [2 marks]

• Plot: point at x = 3 (right), y = -2 (down).
• Quadrant: III (third quadrant, where x > 0 and y < 0 is actually Quadrant IV; correction: x positive, y negative → Quadrant IV).
  Teaching note: Cartesian plane: QI (+,+), QII (−,+), QIII (−,−), QIV (+,−). (3,−2) is (+,−) → QIV.
• Mark: 1 for correct plot, 1 for Quadrant IV.
  Answer: Quadrant IV

Q2 [1 mark]
From diagram, B is at x = −4, y = 1.
Answer: $(-4, 1)$

Q3 [2 marks]
$P(1,2), Q(7,2)$: same y, horizontal distance = $|7 - 1| = 6$.
Answer: 6 units

Q4 [2 marks]
Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 3}{2 - (-1)} = \frac{-6}{3} = -2$.
Answer: $-2$

Q5 [1 mark]
Equation $y = mx + c$, gradient is $m = 4$.
Answer: 4

Section B: Lines and Equations

Q6 [2 marks]
Gradient 2, y-intercept $-3$ → $y = 2x - 3$.
Answer: $y = 2x - 3$

Q7 [3 marks]
(a) $m = \frac{14 - 5}{4 - 1} = \frac{9}{3} = 3$ [1]
(b) $y = 3x + c$; use (1,5): $5 = 3(1) + c \Rightarrow c = 2$. So $y = 3x + 2$ [2]
Answer: (a) 3 (b) $y = 3x + 2$

Q8 [2 marks]
x-axis: $y = 0$. $0 = -x + 6 \Rightarrow x = 6$. Point $T(6, 0)$.
Answer: $(6, 0)$

Q9 [3 marks]
(a) $2x + 1 = -x + 7 \Rightarrow 3x = 6 \Rightarrow x = 2$ [2]
(b) $y = 2(2)+1 = 5$ [1]
Answer: (a) 2 (b) 5

Q10 [2 marks]
Both have gradient 3 → parallel (same gradient, different intercept).
Answer: Parallel, because both lines have gradient 3.

Section C: Quadratic and Other Graphs

Q11 [2 marks]
$x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3)=0 \Rightarrow x=1, 3$.
Answer: $x = 1$ and $x = 3$

Q12 [2 marks]
Min at $x = -\frac{b}{2a} = \frac{6}{2} = 3$. $y = 9 - 18 + 5 = -4$. Point $(3, -4)$.
Answer: $(3, -4)$

Q13 [3 marks]
$y = (x-2)^2$: vertex (2,0); y-intercept at x=0: $y=4$. Sketch upward parabola.
Vertex: (2,0); y-int: (0,4). [3 total: 1 vertex, 1 y-int, 1 sketch]
Answer: Vertex (2,0), y-intercept (0,4)

Q14 [2 marks]
$x=2 \Rightarrow y = 6/2 = 3$. Point (2,3).
Answer: $(2, 3)$

Q15 [3 marks]
(a) $x=1$: $y = 1^2+1=2$; $y = 5-1=4$ → not same. Correction from question intent: use x=2? Actually at x=2: 4+1=5, 5-2=3 no. At x=1, first=2, second=4. The question says show same; if we use x=1 for both: not same. Let us use x=2? No. Let us solve: x^2+1 = 5-x => x^2+x-4=0, roots not integer. The question likely meant x=1 for first only? We follow: (a) at x=1, y=2 for first, y=4 for second – they are not same; but per question we show calculation. We note error in quiz; for answer we compute correctly: intersection when x^2+1=5-x => x≈1.56. But for key: (a) show substitution, (b) one point approx (1.56, 3.44), (c) quadratic and line intersect at 2 points because discriminant >0.
Given artefact, answer key: (a) Sub: $1^2+1=2$, $5-1=4$ (not equal, but demonstrated), (b) solve: $x = \frac{-1+\sqrt{17}}{2} \approx 1.56, y \approx 3.44$, (c) line and parabola meet twice as equation has 2 real roots.
Answer: (a) shown (b) (1.56, 3.44) (c) discriminant positive

Q16 [2 marks]
$C = 3 + 2d$ or $C = 2d + 3$.
Answer: $C = 2d + 3$

Q17 [3 marks]
(a) Plot (0,800),(1,600),(2,400) joined. [1]
(b) Gradient = $\frac{400-800}{2-0} = -200$ [1]
(c) Represents loss of $200 per year. [1] **Answer:** (b) -200 (c) depreciation$200/yr

Q18 [2 marks]
$f(x)+2$ shifts up 2: (1,3)→(1,5).
Answer: (1,5)

Q19 [3 marks]
(a) $t=1: h = -1+4+1 = 4$ [1]
(b) Max at $t = -b/2a = -4/(2(-1)) = 2$, $h = -4+8+1=5$ [2]
Answer: (a) 4 m (b) 5 m

Q20 [2 marks]
$L$ gradient = $8/4=2$. Perpendicular gradient = $-1/2$.
Answer: $-\frac{1}{2}$