Year 9 Mathematics | Victorian Curriculum 2.0
Function transformations
Topic 07 | Number & Algebra | Answer key

Tier 1

    1. (0,7)(0, 7)
    2. (0,−4)(0, -4)
    3. Narrower (since 5>15 > 1).
    4. Wider (since 0.3<10.3 < 1).
    5. Downward.
    6. (6,0)(6, 0)
    7. (−2,−1)(-2, -1)
    8. y=x2−5y = x^2 - 5
    9. y=(x−3)2+4y = (x - 3)^2 + 4

Tier 2

    1. (a) (1,2)(1, 2), (b) (−3,−5)(-3, -5), (c) (0,4)(0, 4).
    2. y=(x−2)2−3y = (x - 2)^2 - 3.
    3. Method 1: vertical translation y=x2+9y = x^2 + 9. Method 2: vertical stretch y=9x2y = 9x^2 (passes through (1,9)(1, 9) but vertex is still at origin, so only the translation gives vertex (0,9)(0, 9)). Accept y=x2+9y = x^2 + 9 and y=−(x)2+9y = -(x)^2 + 9 (reflection plus shift).
    4. a=6a = 6. Method: 6=a×126 = a \times 1^2.
    5. Widest to narrowest: y=14x2y = \tfrac{1}{4}x^2, y=x2y = x^2, y=2x2y = 2x^2, y=4x2y = 4x^2.
    6. One possible answer: y=−2(x−1)2+5y = -2(x - 1)^2 + 5. Accept any equation with a<−1a < -1, h=1h = 1, k=5k = 5.

Tier 3

    1. Replacing xx with (x−h)(x - h) means we need a larger xx-value to produce the same output. For example, the vertex of y=x2y = x^2 is at x=0x = 0; for y=(x−3)2y = (x - 3)^2 the output is 00 when x−3=0x - 3 = 0, i.e. x=3x = 3. Every point shifts right by hh.
    2. The vertex is at (2,0)(2, 0), not (−2,0)(-2, 0). Sam confused the sign: (x−2)(x - 2) means shift right 22.
    3. Maximum height is 88 m, occurring at x=1x = 1 s. The vertex (1,8)(1, 8) gives the peak because a=−5<0a = -5 < 0 (opens downward).
    4. Both have the same shape (a=2a = 2) and the same axis of symmetry (x=3x = 3). The second is a vertical translation of the first, shifted 55 units down. Vertical distance between vertices: 1−(−4)=51 - (-4) = 5 units.

Challenge

    1. y=−2(x−2)2+8y = -2(x - 2)^2 + 8. Method: axis of symmetry at x=0+42=2x = \tfrac{0+4}{2} = 2; vertex (2,8)(2, 8); sub (0,0)(0, 0): 0=a(0−2)2+80 = a(0-2)^2 + 8, so 4a=−84a = -8, a=−2a = -2.
    2. a=1a = 1, h=3h = 3, k=−4k = -4. Method: axis of symmetry at x=0+62=3x = \tfrac{0+6}{2} = 3; minimum is k=−4k = -4; vertex (3,−4)(3, -4); sub (0,5)(0, 5): 5=a(9)−45 = a(9) - 4, 9a=99a = 9, a=1a = 1.
    3. a=−2a = -2, h=3h = 3, k=7k = 7. Method: vertex gives h=3h = 3, k=7k = 7; sub (5,−1)(5, -1): −1=a(5−3)2+7-1 = a(5-3)^2 + 7, 4a=−84a = -8, a=−2a = -2. Opens downward since a<0a < 0.
    4. The axis of symmetry is the midpoint of the roots pp and qq, so x=p+q2x = \tfrac{p+q}{2}. Substituting: y=(p+q2−p)(p+q2−q)=(q−p2)(p−q2)=−(p−q)24y = \bigl(\tfrac{p+q}{2} - p\bigr)\bigl(\tfrac{p+q}{2} - q\bigr) = \bigl(\tfrac{q-p}{2}\bigr)\bigl(\tfrac{p-q}{2}\bigr) = -\tfrac{(p-q)^2}{4}. Vertex: (p+q2,  −(p−q)24)\bigl(\tfrac{p+q}{2},\; -\tfrac{(p-q)^2}{4}\bigr).
Year 9 Mathematics study companion | Answer key