Quadratic Functions — Graphs and Features — Solutions

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1. Identify concavity, y-intercept, axis of symmetry. Fluency

   1. (a) f(x) = x² + 4x − 5 (i) Opens up (a = 1 > 0).
      (ii) y-intercept: f(0) = −5, so (0, −5).
      (iii) Axis of symmetry: x = −4/(2×1) = −2.
   2. (b) g(x) = −2x² + 6x (i) Opens down (a = −2 < 0).
      (ii) y-intercept: g(0) = 0, so (0, 0).
      (iii) Axis of symmetry: x = −6/(2×−2) = 6/4 = 3/2.
   3. (c) h(x) = 3x² − 12x + 1 (i) Opens up (a = 3 > 0).
      (ii) y-intercept: h(0) = 1, so (0, 1).
      (iii) Axis of symmetry: x = −(−12)/(2×3) = 12/6 = 2.
   4. (d) k(x) = −x² + 2x + 8 (i) Opens down (a = −1 < 0).
      (ii) y-intercept: k(0) = 8, so (0, 8).
      (iii) Axis of symmetry: x = −2/(2×−1) = 1.

2. Discriminant and number of x-intercepts. Fluency

   1. (a) y = x² − 5x + 4 Δ = (−5)² − 4(1)(4) = 25 − 16 = 9 > 0. Two x-intercepts.
   2. (b) y = x² − 4x + 4 Δ = (−4)² − 4(1)(4) = 16 − 16 = 0. One x-intercept (tangent to x-axis).
   3. (c) y = 2x² + x + 3 Δ = 1² − 4(2)(3) = 1 − 24 = −23 < 0. No x-intercepts (parabola entirely above x-axis).
   4. (d) y = −x² + 6x − 9 Δ = 6² − 4(−1)(−9) = 36 − 36 = 0. One x-intercept (tangent to x-axis).

3. Find the turning point. Fluency

   1. (a) y = x² − 6x + 2 x = −(−6)/(2) = 3. y = 9 − 18 + 2 = −7. TP: (3, −7), minimum.
   2. (b) y = −x² + 4x + 1 x = −4/(2×−1) = 2. y = −4 + 8 + 1 = 5. TP: (2, 5), maximum.
   3. (c) y = 2x² + 8x − 3 x = −8/(2×2) = −2. y = 2(4) + 8(−2) − 3 = 8 − 16 − 3 = −11. TP: (−2, −11), minimum.
   4. (d) y = −3x² + 12x − 7 x = −12/(2×−3) = 2. y = −3(4) + 24 − 7 = −12 + 24 − 7 = 5. TP: (2, 5), maximum.

4. Vertex form by completing the square. Fluency

   1. (a) f(x) = x² + 6x + 7 f(x) = (x² + 6x + 9) − 9 + 7 = (x + 3)² − 2. TP: (−3, −2), minimum.
   2. (b) f(x) = x² − 4x + 1 f(x) = (x² − 4x + 4) − 4 + 1 = (x − 2)² − 3. TP: (2, −3), minimum.
   3. (c) f(x) = 2x² − 8x + 9 f(x) = 2(x² − 4x) + 9 = 2[(x − 2)² − 4] + 9 = 2(x − 2)² − 8 + 9 = 2(x − 2)² + 1. TP: (2, 1), minimum.
   4. (d) f(x) = −x² + 2x + 5 f(x) = −(x² − 2x) + 5 = −[(x − 1)² − 1] + 5 = −(x − 1)² + 1 + 5 = −(x − 1)² + 6. TP: (1, 6), maximum.

5. All features. Fluency

   1. (a) y = x² + 2x − 8 Axis: x = −2/2 = −1.
      TP: y = 1 − 2 − 8 = −9 → (−1, −9), minimum.
      y-int: (0, −8).
      x-ints: x² + 2x − 8 = 0 → (x+4)(x−2) = 0 → x = −4 or x = 2.
   2. (b) y = −x² + 5x − 4 Axis: x = −5/(2×−1) = 5/2.
      TP: y = −25/4 + 25/2 − 4 = −25/4 + 50/4 − 16/4 = 9/4 → (5/2, 9/4), maximum.
      y-int: (0, −4).
      x-ints: −x² + 5x − 4 = 0 → x² − 5x + 4 = 0 → (x−1)(x−4) = 0 → x = 1 or x = 4.
   3. (c) y = x² − 4x + 5 Axis: x = 4/2 = 2.
      TP: y = 4 − 8 + 5 = 1 → (2, 1), minimum.
      y-int: (0, 5).
      Δ = 16 − 20 = −4 < 0: no x-intercepts. Parabola entirely above x-axis.

6. Finding the equation. Understanding

   1. (a) x-intercepts −2 and 4, passes through (0, −16). y = a(x + 2)(x − 4). Substitute (0, −16): −16 = a(2)(−4) = −8a → a = 2.
      y = 2(x + 2)(x − 4) = 2(x² − 2x − 8) = 2x² − 4x − 16.
   2. (b) Vertex (3, −2), passes through (5, 6). Vertex form: y = a(x − 3)² − 2. Substitute (5, 6): 6 = a(4) − 2 → 8 = 4a → a = 2.
      Vertex form: y = 2(x − 3)² − 2.
      Standard form: y = 2(x² − 6x + 9) − 2 = 2x² − 12x + 18 − 2 = 2x² − 12x + 16.
   3. (c) Opens downward, axis x = 1, passes through (0, 3). y = a(x − 1)² + k, opens down so a < 0.
      Passes through (0, 3): 3 = a(1) + k → k = 3 − a. …(1)
      Passes through (2, 3) (symmetric about x=1): same equation gives same result — consistent.
      Actually we need one more condition. Since (0,3) and (2,3) are both on the curve and symmetric about x=1, they give the same equation, so we can only determine k − a = 3, i.e., k = a + 3.
      Wait: substitute (0, 3): 3 = a(0−1)² + k = a + k, so a + k = 3.
      The maximum value is k (at the vertex). We need another piece of info to pin down a and k individually. However, the question only asks for the maximum value, which is k.
      From (0, 3): a + k = 3, and since both (0, 3) and (2, 3) give the same equation (symmetry is consistent), we cannot determine a uniquely without more data. If we assume the parabola also touches the x-axis: Δ = 0, but that's not given.
      With only the given info: the maximum value k = 3 − a. The maximum is not uniquely determined without knowing a. — Note: This question requires an additional constraint to solve fully (e.g., a specific x-intercept). As stated, k = 3 − a for any a < 0.

7. Transformations of parabolas. Understanding

   1. (a) y = (x + 3)² − 1 Translation 3 units left and 1 unit down. TP: (−3, −1), minimum.
   2. (b) y = −2(x − 1)² + 4 Reflect in x-axis (a < 0), vertically stretch by factor 2, translate 1 right and 4 up. TP: (1, 4), maximum.
   3. (c) y = (x − 2)² Translation 2 units right. TP: (2, 0), minimum (touches x-axis).
      Expand: y = x² − 4x + 4. y-intercept: (0, 4).

8. Range and practical contexts. Understanding

   1. (a) Range of f(x) = (x − 2)² + 3. Minimum value is 3 (at x = 2). Range: [3, ∞).
   2. (b) Range of g(x) = −x² + 4x − 1. Complete the square: g(x) = −(x − 2)² + 4 − 1 = −(x − 2)² + 3. Maximum value is 3. Range: (−∞, 3].
   3. (c) h(t) = −5t² + 20t + 1. Axis: t = −20/(2×−5) = 2 seconds.
      h(2) = −5(4) + 40 + 1 = −20 + 40 + 1 = 21 m.
      Maximum height: 21 m at t = 2 s.

9. Intersection of a parabola and a line. Problem Solving

   1. (a) Features of P: y = x² − 3x + 2. Axis: x = 3/2. TP: y = 9/4 − 9/2 + 2 = 9/4 − 18/4 + 8/4 = −1/4 → (3/2, −1/4), minimum.
      y-int: (0, 2).
      x-ints: (x−1)(x−2) = 0 → x = 1 or x = 2.
   2. (b) Points of intersection of P and L. Set equal: x² − 3x + 2 = x − 1.
      x² − 4x + 3 = 0 → (x−1)(x−3) = 0 → x = 1 or x = 3.
      y = 1 − 1 = 0 or y = 3 − 1 = 2.
      Intersection points: (1, 0) and (3, 2).
   3. (c) Line y = x + k tangent to P (one intersection). x² − 3x + 2 = x + k → x² − 4x + (2 − k) = 0.
      For one solution: Δ = 0 → 16 − 4(2 − k) = 0 → 16 − 8 + 4k = 0 → 8 + 4k = 0 → k = −2.
      The line y = x − 2 is tangent to the parabola — it touches at exactly one point without crossing.

10. Discriminant conditions. Problem Solving

    1. (a) Discriminant of f(x) = x² + kx + (k + 3). Δ = k² − 4(1)(k + 3) = k² − 4k − 12.
    2. (b) No x-intercepts (Δ < 0). k² − 4k − 12 < 0.
       Factorise: (k − 6)(k + 2) < 0.
       Sign analysis: parabola in k opens up, roots at k = −2 and k = 6.
       Δ < 0 for −2 < k < 6.
    3. (c) Just touches x-axis (Δ = 0). (k − 6)(k + 2) = 0 → k = 6 or k = −2.
       For k = 6: f(x) = x² + 6x + 9 = (x+3)². TP: (−3, 0).
       For k = −2: f(x) = x² − 2x + 1 = (x−1)². TP: (1, 0).
    4. (d) k = −4: f(x) = x² − 4x − 1 in vertex form. f(x) = (x² − 4x + 4) − 4 − 1 = (x − 2)² − 5.
       TP: (2, −5), minimum. Axis: x = 2. y-int: (0, −1).
       Δ = 16 + 4 = 20 > 0: two x-intercepts at x = (4 ± √20)/2 = 2 ± √5.