Secondary 4 Elementary Mathematics Algebra Functions Quiz

Secondary 4 Elementary Mathematics Quiz - Algebra Functions

Answer Key

Question 1 [3 marks]

(a) Vertex = (3, 2) [1]

The function is in the form y = (x − p)² + q, so the vertex is (p, q) = (3, 2).

(b) Minimum [1]

Since the coefficient of (x − 3)² is positive (+1), the parabola opens upwards, so the vertex is a minimum point.

(c) x = 3 [1]

The axis of symmetry passes through the vertex, so x = 3.

Question 2 [5 marks]

(a) x-intercepts: x = 2 and x = −6 [2]

Set y = 0: (x − 2)(x + 6) = 0, so x = 2 or x = −6.

(b) y-intercept: y = −12 [1]

Set x = 0: y = (0 − 2)(0 + 6) = (−2)(6) = −12.

(c) Vertex = (−2, −16) [2]

The x-coordinate of the vertex is the midpoint of the x-intercepts: x = (2 + (−6))/2 = −4/2 = −2. Substitute x = −2: y = (−2 − 2)(−2 + 6) = (−4)(4) = −16. So vertex = (−2, −16).

Question 3 [6 marks]

(a) Vertex = (−1, 9) [1]

The function is in the form y = −(x − p)² + q where p = −1 and q = 9.

(b) x-intercepts: x = −4 and x = 2 [2]

Set y = 0: −(x + 1)² + 9 = 0 (x + 1)² = 9 x + 1 = ±3 x = −1 ± 3 x = 2 or x = −4.

(c) Sketch [3]

• Parabola opening downwards (coefficient is negative)
• Vertex at (−1, 9)
• x-intercepts at (−4, 0) and (2, 0)
• y-intercept: set x = 0: y = −(0 + 1)² + 9 = −1 + 9 = 8, so (0, 8)
• Award 1 mark for correct shape, 1 mark for correct vertex, 1 mark for correct intercepts.

Question 4 [5 marks]

(a) y = (x − 2)² − 9 [3]

x² − 4x − 5 = x² − 4x + 4 − 4 − 5 = (x − 2)² − 9

Award 1 mark for (x − 2)², 1 mark for correct constant, 1 mark for complete expression.

(b) Vertex = (2, −9) [1]

(c) y-intercept: y = −5 [1]

Set x = 0: y = 0² − 4(0) − 5 = −5.

Question 5 [5 marks]

(a) a = −2, b = 3 (or vice versa) [2]

The x-intercepts are x = −2 and x = 3, so the factors are (x + 2) and (x − 3), meaning a = −2 and b = 3.

(b) Vertex = (0.5, −6.25) [2]

x-coordinate of vertex = (−2 + 3)/2 = 0.5. Substitute x = 0.5: y = (0.5 + 2)(0.5 − 3) = (2.5)(−2.5) = −6.25.

(c) x = 0.5 [1]

Question 6 [4 marks]

(a) a = 3 [2]

Substitute (2, 24): 24 = a(2)³ = 8a, so a = 24/8 = 3.

(b) y = −3 [1]

y = 3(−1)³ = 3(−1) = −3.

(c) n = 3 [1]

Question 7 [4 marks]

(a) k = 20 [2]

Substitute (2, 5): 5 = k/(2)² = k/4, so k = 20.

(b) y = 1.25 [1]

y = 20/(4)² = 20/16 = 5/4 = 1.25.

(c) y decreases / y approaches zero [1]

As x increases, x² increases, so k/x² decreases towards zero.

Question 8 [5 marks]

(a) Table [2]

Award 1 mark for each correct pair (or 1 mark for 3+ correct, 0 for fewer).

(b) Sketch [2]

• Curve passing through (−1, 1/3), (0, 1), (1, 3), (2, 9)
• Increasing exponential curve
• Approaches y = 0 as x decreases (horizontal asymptote)
• Award 1 mark for correct shape, 1 mark for correct points.

(c) y = 0 [1]

Question 9 [5 marks]

(a) y = 2 [1]

y = 2 × 5⁰ = 2 × 1 = 2.

(b) y = 50 [1]

y = 2 × 5² = 2 × 25 = 50.

(c) x = 3 [2]

250 = 2 × 5ˣ 125 = 5ˣ 5³ = 5ˣ x = 3.

Award 1 mark for 125 = 5ˣ, 1 mark for x = 3.

(d) Growth [1]

Since the base 5 > 1, the function represents exponential growth.

Question 10 [4 marks]

(a) y = 4ˣ is steeper [2]

For x > 0, 4ˣ grows faster than 2ˣ because 4 > 2. The base of the exponential is larger, so the rate of increase is greater.

Award 1 mark for identifying y = 4ˣ, 1 mark for correct reasoning.

(b) (0, 1) [1]

4⁰ = 1 and 2⁰ = 1, so both pass through (0, 1).

(c) y > 0 [1]

The range of an exponential function y = aˣ (where a > 0) is all positive real numbers.

Question 11 [4 marks]

(a) Tangent drawn at (2, 4) [1]

A straight line touching the curve at exactly one point (2, 4) with approximately the correct gradient.

(b) Gradient ≈ 4 [2]

The gradient of y = x² at x = 2 is dy/dx = 2x = 4. Students should draw a tangent and estimate; accept answers in the range 3.5 to 4.5.

(c) Depends on tangent drawn; accept overestimate or underestimate with valid reason [1]

If the tangent is drawn slightly above the true tangent, the gradient will be an overestimate. If below, an underestimate.

Question 12 [4 marks]

(a) Gradient = 6 [2]

When x = 1: y = 1² + 2(1) = 3. Point = (1, 3). When x = 3: y = 3² + 2(3) = 15. Point = (3, 15). Gradient = (15 − 3)/(3 − 1) = 12/2 = 6.

(b) Draw a tangent to the curve at the point where x = 2, then calculate the gradient of that tangent [2]

The gradient of the chord gives an approximation. To estimate the gradient at x = 2, draw a tangent at that point and find its gradient using two points on the tangent line.

Question 13 [5 marks]

(a) 17 m/s [2]

When t = 1: d = 3(1)² + 5(1) = 8. When t = 3: d = 3(9) + 15 = 42. Average rate = (42 − 8)/(3 − 1) = 34/2 = 17 m/s.

(b) Gradient ≈ 17 m/s [2]

The instantaneous rate at t = 2 should be close to the average rate between t = 1 and t = 3. Students draw a tangent at t = 2 and estimate; accept 16 to 18.

(c) metres per second (m/s) [1]

Question 14 [5 marks]

(a) 5 m [1]

Set t = 0: h = −2(0)² + 12(0) + 5 = 5.

(b) Vertex = (3, 23) [3]

t = −b/(2a) = −12/(2 × −2) = −12/−4 = 3. h = −2(9) + 12(3) + 5 = −18 + 36 + 5 = 23. The vertex represents the maximum height of the ball, which occurs at t = 3 seconds.

Award 1 mark for t = 3, 1 mark for h = 23, 1 mark for interpretation.

(c) 23 m [1]

Question 15 [5 marks]

(a) A = 20x − x² [2]

Perimeter = 40, so 2x + 2w = 40, giving w = 20 − x. Area = x × w = x(20 − x) = 20x − x².

(b) A = −(x − 10)² + 100 [2]

20x − x² = −(x² − 20x) = −(x² − 20x + 100 − 100) = −(x − 10)² + 100.

(c) 100 m² [1]

Maximum area occurs at the vertex: when x = 10, A = 100.

Question 16 [5 marks]

(a) f(0) = 5 [1]

f(0) = 0² − 6(0) + 5 = 5.

(b) x = 1 and x = 5 [2]

x² − 6x + 5 = 0 (x − 1)(x − 5) = 0 x = 1 or x = 5.

(c) f(x) = (x − 3)² − 4 [2]

x² − 6x + 5 = x² − 6x + 9 − 9 + 5 = (x − 3)² − 4.

Question 17 [5 marks]

(a) g(0) = 2 [1]

g(0) = 2⁰⁺¹ = 2¹ = 2.

(b) g(3) = 16 [1]

g(3) = 2³⁺¹ = 2⁴ = 16.

(c) x = 4 [2]

2ˣ⁺¹ = 32 = 2⁵ x + 1 = 5 x = 4.

(d) g(x) = 2 × 2ˣ, k = 2 [1]

2ˣ⁺¹ = 2ˣ × 2¹ = 2 × 2ˣ.

Question 18 [6 marks]

(a) (x − 4)(x + 2) [2]

x² − 2x − 8 = (x − 4)(x + 2).

(b) x-intercepts: x = 4 and x = −2; y-intercept: y = −8 [2]

Set y = 0: (x − 4)(x + 2) = 0, so x = 4 or x = −2. Set x = 0: y = −8.

(c) Vertex = (1, −9) [2]

x-coordinate = (4 + (−2))/2 = 1. y = (1)² − 2(1) − 8 = 1 − 2 − 8 = −9.

Question 19 [5 marks]

(a) a = 12 [2]

Substitute (4, 3): 3 = a/4, so a = 12.

(b) y = 2 [1]

y = 12/6 = 2.

(c) x = 0 and y = 0 [2]

The graph of y = a/x has vertical asymptote x = 0 (y-axis) and horizontal asymptote y = 0 (x-axis).

Question 20 [5 marks]

(a) y = k(x + 1)(x − 5) [1]

The x-intercepts are −1 and 5, so the factors are (x + 1) and (x − 5).

(b) k = 1 [2]

Substitute (2, −9): −9 = k(2 + 1)(2 − 5) = k(3)(−3) = −9k. So k = 1.

(c) Vertex = (2, −9) [2]

x-coordinate = (−1 + 5)/2 = 2. Substitute x = 2: y = (2 + 1)(2 − 5) = (3)(−3) = −9. Vertex = (2, −9).

Mark Summary

Note: The total marks sum to 90. If the quiz is to be out of 50, scale marks proportionally or select a subset of questions. The mark allocations above reflect the difficulty and working required for each question.

Common Mistakes to Watch For

• Completing the square: Forgetting to subtract the added constant (e.g., adding 4 but not subtracting 4).
• Vertex from intercepts: Using the wrong formula for the x-coordinate of the vertex; it is the midpoint of the x-intercepts.
• Exponential equations: Not recognising that 32 = 2⁵ or 125 = 5³.
• Gradient of a curve: Confusing the gradient of a chord with the gradient of a tangent.
• Sign errors: When expanding −(x + 1)², students often write −x² + 1 instead of −x² − 2x − 1.
• Asymptotes: Forgetting that y = a/x has both x = 0 and y = 0 as asymptotes.