Year 10 Mathematics | Victorian Curriculum 2.0
Non-linear graphs
Topic 06 | Number & Algebra | Answer key

Year 10 core - answers

Fluency

Tier 1: basic skills

    1. Vertex (−1,−9)(-1, -9). Opens upward (a=1>0a = 1 > 0).
    2. x2−4x−5=(x−5)(x+1)=0x^2 - 4x - 5 = (x - 5)(x + 1) = 0. xx-intercepts: x=5x = 5 and x=−1x = -1.
    3. Centre (3,−2)(3, -2), radius 55.
    4. x2+(y−4)2=36x^2 + (y - 4)^2 = 36.
    5. x=0x = 0: y=1y = 1. x=2x = 2: y=9y = 9. x=−1x = -1: y=13y = \tfrac{1}{3}.
    6. x=1x = 1: y=4y = 4. x=4x = 4: y=1y = 1. x=−2x = -2: y=−2y = -2.
    7. (a) Exponential. (b) Circle. (c) Hyperbola. (d) Parabola.
    8. Horizontal asymptote: y=0y = 0. Vertical asymptote: x=0x = 0.
Reasoning

Tier 2: mixed practice

    1. y=(x2+8x+16)−16+12=(x+4)2−4y = (x^2 + 8x + 16) - 16 + 12 = (x + 4)^2 - 4. Vertex (−4,−4)(-4, -4).
    2. t=−19.62(−4.9)=2t = -\dfrac{19.6}{2(-4.9)} = 2 seconds. h=−4.9(4)+19.6(2)+1.5=−19.6+39.2+1.5=21.1h = -4.9(4) + 19.6(2) + 1.5 = -19.6 + 39.2 + 1.5 = 21.1 metres.
    3. Substitute: (3−1)2+(4−2)2=4+4=8(3-1)^2 + (4-2)^2 = 4 + 4 = 8. Since 8<9=r28 < 9 = r^2, the point is inside the circle.
    4. y=(12)x=2−xy = \left(\tfrac{1}{2}\right)^x = 2^{-x}, so the second curve is a reflection of y=2xy = 2^x in the yy-axis. Both pass through (0,1)(0, 1). One grows right, the other decays right.
    5. 5=k25 = \tfrac{k}{2}, so k=10k = 10. Asymptotes: x=0x = 0 and y=0y = 0.
    6. 11=a(3−1)2+3=4a+311 = a(3-1)^2 + 3 = 4a + 3. 4a=84a = 8. a=2a = 2.
Reasoning

Tier 3: explain and apply

    1. The base goes from (−4,0)(-4, 0) to (4,0)(4, 0). Maximum height at (0,5)(0, 5). Equation: y=ax2+5y = a x^2 + 5. At x=4x = 4: 0=16a+50 = 16a + 5, so a=−516a = -\tfrac{5}{16}. Equation: y=−516x2+5y = -\tfrac{5}{16}x^2 + 5. The truck is 33 m wide, so its edges are at x=±1.5x = \pm 1.5. Height at x=1.5x = 1.5: y=−516(2.25)+5=−0.703+5=4.297y = -\tfrac{5}{16}(2.25) + 5 = -0.703 + 5 = 4.297 m. Since 4.297>44.297 > 4, yes, the truck can pass through.
    2. Substitute y=x+1y = x + 1 into x2+y2=25x^2 + y^2 = 25: x2+(x+1)2=25x^2 + (x+1)^2 = 25. 2x2+2x+1=252x^2 + 2x + 1 = 25. 2x2+2x−24=02x^2 + 2x - 24 = 0. x2+x−12=0x^2 + x - 12 = 0. (x+4)(x−3)=0(x+4)(x-3) = 0. x=−4x = -4 or x=3x = 3. Points: (−4,−3)(-4, -3) and (3,4)(3, 4).
    3. y=3×2xy = 3 \times 2^x has the same shape as y=2xy = 2^x but every yy-value is multiplied by 33. It is a vertical stretch by factor 33. The yy-intercept moves from (0,1)(0, 1) to (0,3)(0, 3). The asymptote remains y=0y = 0.
    4. For y=kxy = \tfrac{k}{x} to equal zero, we would need kx=0\tfrac{k}{x} = 0, which means k=0×x=0k = 0 \times x = 0. But k≠0k \neq 0 (otherwise it is not a hyperbola), so no value of xx can make y=0y = 0. As xx grows larger, yy gets closer and closer to 00 but never reaches it.
Reasoning

Challenge

    1. Vertex form: y=a(x−2)2−3y = a(x - 2)^2 - 3. Through (0,5)(0, 5): 5=a(4)−35 = a(4) - 3, a=2a = 2. Expand: y=2(x2−4x+4)−3=2x2−8x+5y = 2(x^2 - 4x + 4) - 3 = 2x^2 - 8x + 5. So a=2a = 2, b=−8b = -8, c=5c = 5.
    2. From y=x2y = x^2, substitute into x2+y2=2x^2 + y^2 = 2: y+y2=2y + y^2 = 2 (since x2=yx^2 = y). y2+y−2=0y^2 + y - 2 = 0. (y+2)(y−1)=0(y+2)(y-1) = 0. y=1y = 1 (take y≥0y \geq 0). Then x2=1x^2 = 1, so x=±1x = \pm 1. Points: (1,1)(1, 1) and (−1,1)(-1, 1).
    3. Reflect in yy-axis: y=2−xy = 2^{-x}. Shift up 33: y=2−x+3y = 2^{-x} + 3. Asymptote: y=3y = 3.
Year 10 Mathematics study companion | Answer key