NCEA Level 2ModellingAS 91261

Algebraic modelling: 6 exam questions solved step by step

What this lesson covers

  • A model is completed by substituting known facts from the context into its given form.
  • Every answer from a model has to be checked against the context it came from.
Key idea

6 models, one routine

  • 1. Write the form the context gives.
  • 2. Substitute known facts to finish it.
  • 3. Solve: get to zero, or use ±√.
  • 4. Keep only answers that fit.
  • 5. Answer in words, with units.
Your turn

Question 1 · Achieved, in context

  • A fountain jet lands 4 m from its nozzle.
  • Height: h = kx(4 − x) m, x m along.
  • At x = 1 m the jet is 1.5 m high.
  • Find k, then the greatest height.
Worked example

Q1: complete the model

  • Substitute the known point.
  • Top: halfway between roots.
  1. h = kx(4 − x)zero at x = 0 and 4
  2. 1.5 = k × 1 × (4 − 1)x = 1, h = 1.5
  3. 1.5 = 3k
  4. k = 0.5÷ 3
  5. h = 0.5x(4 − x)
  6. top at x = 2halfway between roots
  7. h = 0.5 × 2 × (4 − 2)
  8. = 0.5 × 2 × 2 = 2
  9. The jet is at most 2 m high.
Your turn

Question 2 · Achieved, in context

  • A rugby ball is kicked from the ground.
  • It is highest, 16 m up, at x = 10 m.
  • Model: y = a(x − 10)² + 16
  • Find a, then the height at x = 15 m.
Worked example

Q2: vertex form

  • The kick starts at (0, 0).
  • Square before × a.
  1. y = a(x − 10)² + 16top at (10, 16)
  2. 0 = a(0 − 10)² + 16starts at (0, 0)
  3. 0 = 100a + 16
  4. 100a = −16−16 both sides
  5. a = −0.16÷ 100
  6. y = −0.16(x − 10)² + 16
  7. y = −0.16(15 − 10)² + 16x = 15
  8. = −0.16 × 25 + 16 = −4 + 16
  9. = 12: the ball is 12 m high
Your turn

Question 3 · Merit, in context

  • A lawn 10 m by 6 m has a path x m wide
  • all the way round it.
  • Lawn and path together cover 96 m².
  • Find the width of the path.
  • Careful: the path is on both sides.
Worked example

Q3: an area model

  • Add 2x to each side.
  • A width cannot be negative.
  1. (10 + 2x) by (6 + 2x)path on every side
  2. (10 + 2x)(6 + 2x) = 96area
  3. 60 + 20x + 12x + 4x² = 96expand
  4. 4x² + 32x − 36 = 0collect, −96
  5. x² + 8x − 9 = 0÷ 4
  6. (x + 9)(x − 1) = 09 × −1, 9 + −1
  7. x = −9a width is positive
  8. x = 1: 12 × 8 = 96
  9. The path is 1 m wide.
Your turn

Question 4 · Merit, in context

  • The kick from question 2:
  • y = −0.16(x − 10)² + 16
  • How far away is the ball 10 m high?
  • Give answers to 1 decimal place.
Worked example

Q4: solve the finished model

  • Undo in reverse order.
  • Keep ±. Round at the end.
  1. −0.16(x − 10)² + 16 = 10y = 10
  2. −0.16(x − 10)² = −6−16 both sides
  3. (x − 10)² = 37.5÷ −0.16
  4. x − 10 = ±√37.5square root: ±
  5. = ±6.1237…
  6. x = 10 − 6.1237 = 3.8763
  7. x = 10 + 6.1237 = 16.1237
  8. both before it lands at x = 20
  9. 3.9 m (up) and 16.1 m (down)
Your turn

Question 5 · Excellence, in context

  • A tunnel mouth is a parabola 8 m wide
  • at the road and 6 m high in the middle.
  • Model: y = ax² + 6, x m from the centre.
  • Can a truck 3 m wide and 4.5 m tall fit,
  • driving down the middle? Pause and try it.
Worked example

Q5: complete the model

  • The road meets it at x = ±4.
  • Use one of those points.
  1. y = ax² + 6top at (0, 6)
  2. 0 = a(4)² + 6meets road at x = 4
  3. 0 = 16a + 6
  4. 16a = −6−6 both sides
  5. a = −0.375÷ 16
  6. y = 6 − 0.375x²
Worked example

Q5: test the tightest point

  • Not the centre: the corner.
  • Corner is 1.5 m out.
  1. y = 6 − 0.375x²
  2. corner at x = 1.5half of 3 m
  3. y = 6 − 0.375 × 1.5²height there
  4. = 6 − 0.375 × 2.25
  5. = 6 − 0.84375 = 5.15625
  6. 5.16 m > 4.5 mroof above the corner
  7. 5.15625 − 4.5 = 0.65625
  8. Yes: about 0.66 m to spare
Your turn

Question 6 · Excellence, in context

  • A garden bed is a right-angled triangle.
  • Its longest side is 10 m.
  • Its area is 24 m².
  • Find the two shorter sides.
Question 6 · Excellence, in context
Worked example

Q6: two facts, one equation

  • Pythagoras and area.
  • Substitute for b.
  1. Let the shorter sides be a and b
  2. a² + b² = 100Pythagoras
  3. ½ab = 24, so ab = 48area
  4. b = 48/a÷ a
  5. a² + (48/a)² = 100substitute
  6. a² + 2304/a² = 100square
  7. a⁴ + 2304 = 100a²× a²
  8. a⁴ − 100a² + 2304 = 0rearrange to zero
Worked example

Q6: a quadratic in a²

  • Let u = a².
  • Reject negative lengths.
  1. a⁴ − 100a² + 2304 = 0
  2. u² − 100u + 2304 = 0let u = a²
  3. (u − 36)(u − 64) = 036 × 64, 36 + 64
  4. a² = 36 or a² = 64u = a²
  5. a = ±6 or a = ±8square root
  6. a = −6, a = −8a length is positive
  7. a = 6, b = 8 or a = 8, b = 6
  8. 36 + 64 = 100, ½ × 6 × 8 = 24
  9. The shorter sides are 6 m and 8 m.
Summary

The routine that solved all six

  • Form, complete, solve, reject with a reason, answer.
StepHow
Rootsy = kx(w − x), then a point
Topy = a(x − h)² + k, a point
ShapesPythagoras, area formulae
Solveto zero, factorise or ±√
Rejectonly with a reason
Answera sentence, with units

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