NCEA Level 2ModellingAS 91261

Algebraic modelling: completing a model from the context

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

Completing a model

  • The exam often gives the shape of the model.
  • Your job: find the missing constant.
  • Then use the model to answer the question.
Worked example

A bridge arch: 20 m wide, 6 m high

  • Put the origin at one foot of the arch.
  • It is zero at 0 and 20: y = kx(20 − x).
  • The top is at x = 10, height 6.
A bridge arch: 20 m wide, 6 m high
Worked example

Finding k

  • Substitute the top point.
  • Then solve for k.
  1. y = kx(20 − x)zero at 0 and 20
  2. 6 = k × 10 × (20 − 10)top: x = 10, y = 6
  3. 6 = 100k
  4. k = 6 ÷ 100 = 0.06÷ 100
  5. y = 0.06x(20 − x)
Worked example

How wide is the arch 4 m up?

  • Set the height to 4 and solve.
  • x = 4.23 m and x = 15.77 m.
  • The width at 4 m is 11.55 m.
How wide is the arch 4 m up?
Worked example

Height 4 m: rearrange to zero

  • Expand the bracket.
  • Clear the decimals.
  1. 0.06x(20 − x) = 4height 4 m
  2. 1.2x − 0.06x² = 4expand
  3. 0.06x² − 1.2x + 4 = 0rearrange to zero
  4. 3x² − 60x + 200 = 0× 50
Worked example

The quadratic formula

  • Two roots: one each side.
  • Width: their difference.
  1. 3x² − 60x + 200 = 0
  2. a = 3, b = −60, c = 200
  3. x = (60 ± √(3600 − 2400))/6formula
  4. = (60 ± √1200)/6
  5. = 10 ± 5.774
  6. x = 4.226 or x = 15.774
  7. width = 15.774 − 4.226difference
  8. ≈ 11.55 m
Worked example

A thrown ball: vertex form

  • Top known: vertex form.
  • One more point finds a.
  1. y = a(x − 5)² + 8top at (5, 8)
  2. 3 = a(0 − 5)² + 8through (0, 3)
  3. 3 = 25a + 8
  4. 25a = −5subtract 8
  5. a = −0.2÷ 25
  6. y = −0.2(x − 5)² + 8
Worked example

Where does the ball land?

  • It lands where y = 0.
  • Reject the negative root.
  1. y = −0.2(x − 5)² + 8
  2. −0.2(x − 5)² + 8 = 0lands: y = 0
  3. −0.2(x − 5)² = −8subtract 8
  4. (x − 5)² = 40÷ −0.2
  5. x − 5 = ±√40 = ±6.325square root
  6. x = 5 − 6.325 = −1.325behind the thrower
  7. x = 5 + 6.325 = 11.325
  8. It lands 11.3 m away.
Key idea

Which form to use

  • Know where it crosses zero? Use the roots.
  • Know the top or bottom? Use vertex form.
  • Then one more point finds the constant.
Worked example

A model from Pythagoras

  • Legs x and x + 7.
  • Hypotenuse 13.
  • A quadratic follows.
  1. x² + (x + 7)² = 13²Pythagoras
  2. x² + x² + 14x + 49 = 169expand
  3. 2x² + 14x − 120 = 0subtract 169
  4. x² + 7x − 60 = 0÷ 2
  5. (x + 12)(x − 5) = 0factorise
  6. x = −12a length is positive
  7. x = 5: legs 5 and 12
  8. area = ½ × 5 × 12 = 30 units²
Key idea

Check every answer against the context

  • x = −12 fits the equation, not the triangle.
  • The ball does not land behind the thrower.
  • Say why each rejected answer fails.
Key idea

Where the marks are

  • Merit: complete the model and use it.
  • Excellence: model the situation yourself.
  • Both: an answer in context, with units.
Summary

What to take away

  • Finish the model from the facts, then use it.
Given a formsubstitute known facts to finish it
Zeros knowny = kx(w − x)
Top knowny = a(x − h)² + k
ShapesPythagoras, area and volume formulae
Solvethen reject what cannot happen
Answera sentence with units
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Sketch the situation first.
  • Answers with units.
Question
1Arch: 12 m wide, 9 m high, y = kx(12 − x). Find k.
2How high is that arch 2 m from one end?
3Legs x and x + 2, hypotenuse 10. Find the legs.
4Q3 gives x = 6 or −8. Which do you keep, and why?
Show answer

1. At x = 6: 9 = 36k, so k = 0.25. 2. y = 0.25 × 2 × 10 = 5 m. 3. x² + (x + 2)² = 100 gives x² + 2x − 48 = 0, so x = 6: legs 6 and 8. 4. Keep 6: a length cannot be negative.

Worked example

The answers: questions 1 and 2

  1. 1. top at x = 6, y = 9halfway across
  2. 9 = k × 6 × (12 − 6)substitute
  3. 9 = 36k
  4. k = 9 ÷ 36 = 0.25
  5. 2. y = 0.25 × 2 × (12 − 2)x = 2
  6. = 0.25 × 2 × 10
  7. = 5 m high
Worked example

The answers: questions 3 and 4

  1. 3. x² + (x + 2)² = 10²Pythagoras
  2. x² + x² + 4x + 4 = 100expand
  3. 2x² + 4x − 96 = 0subtract 100
  4. x² + 2x − 48 = 0÷ 2
  5. (x + 8)(x − 6) = 0factorise
  6. x = 6 or x = −8
  7. legs 6 and 6 + 2 = 8check: 36 + 64 = 100
  8. 4. x = −8 is negativenot a length
  9. keep x = 6

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