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Test solutions Grade 9 : Quadratic Functions and Graphs — Zyro the alien explorer of Planète Maths

Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

1 Features of a parabola / 3 pts

  1. \(a = 1\), \(b = -8\), \(c = 12\); opens up because \(a > 0\). (1 pt)
  2. \(x = -\dfrac{-8}{2} = 4\). (1 pt)
  3. \(y = 16 - 32 + 12 = -4\): vertex \((4, -4)\), minimum value \(-4\). (1 pt)

2 Solving by factoring / 4 pts

  1. Numbers with product \(-35\) and sum 2: 7 and \(-5\). \((x + 7)(x - 5) = 0\) (1 pt), so \(x = -7\) or \(x = 5\) (1 pt).
  2. Numbers with product 24 and sum \(-11\): \(-3\) and \(-8\). \((x - 3)(x - 8) = 0\) (1 pt), so \(x = 3\) or \(x = 8\) (1 pt).

3 Reading a vertex form / 3 pts

  1. Vertex \((3, 18)\); \(a = -2 < 0\), so it opens down. (1 pt)
  2. \(f(0) = -2(9) + 18 = 0\): the \(y\)-intercept is \((0, 0)\). (1 pt)
  3. \(-2(x - 3)^2 + 18 = 0\) gives \((x - 3)^2 = 9\), so \(x - 3 = \pm 3\) and \(x = 0\) or \(x = 6\). (1 pt)

4 Transformations and equations / 4 pts

  1. Shift left 2 and down 3, and shrink vertically by a factor of \(\tfrac{1}{2}\) (wider). (1 pt) The vertex is \((-2, -3)\). (1 pt)
  2. \(y = a(x - 1)^2 + 4\). Substitute \((3, 0)\): \(0 = 4a + 4\), so \(a = -1\) (1 pt). The equation is \(y = -(x - 1)^2 + 4\) (1 pt).

5 Reading a graph / 3 pts

  1. \(A(-1, 0)\), \(B(5, 0)\), \(V(2, 9)\), \(C(0, 5)\) (1 pt). The axis of symmetry is \(x = 2\), halfway between \(A\) and \(B\).
  2. The parabola opens down and passes through \(C\): \(5 = a(1)(-5)\), so \(a = -1\) (1 pt). Then \(y = -(x + 1)(x - 5) = -x^2 + 4x + 5\) (1 pt).

6 A stone off a bridge / 3 pts

  1. \(t = -\dfrac{16}{2(-16)} = 0.5\) second. (1 pt)
  2. \(h(0.5) = -4 + 8 + 20 = 24\) feet. (1 pt)
  3. Solve \(-16t^2 + 16t + 20 = 0\), that is \(4t^2 - 4t - 5 = 0\) (0.5 pt). \(t = \dfrac{4 + \sqrt{96}}{8} \approx 1.72\) seconds; the negative root is rejected (0.5 pt).
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