Solutions — Graphing Parabolas

1. Key features from standard form. Fluency

   • (a) y = x² − 4x + 3: a = 1 > 0 ⇒ opens up. y-intercept: (0, 3). Axis: x = 4/2 = x = 2. Vertex: y(2) = 4−8+3 = −1 ⇒ (2, −1).
   • (b) y = −2x² + 8x − 5: a = −2 < 0 ⇒ opens down. y-intercept: (0, −5). Axis: x = −8/(2×−2) = −8/−4 = x = 2. Vertex: y(2) = −8+16−5 = 3 ⇒ (2, 3).
   • (c) y = x² + 6x: a = 1 > 0 ⇒ opens up. y-intercept: (0, 0). Axis: x = −6/2 = x = −3. Vertex: y(−3) = 9−18 = −9 ⇒ (−3, −9).
   • (d) y = −x² + 4: a = −1 < 0 ⇒ opens down. y-intercept: (0, 4). Axis: x = 0/(2×−1) = x = 0 (the y-axis). Vertex: y(0) = 4 ⇒ (0, 4).

2. x-intercepts. Fluency

   • (a) y = x² − 5x + 6: Set y = 0: x²−5x+6=0. Factorise: (x−2)(x−3)=0. x-intercepts: (2, 0) and (3, 0).
   • (b) y = x² − 4x + 4: Set y = 0: x²−4x+4=0. Factorise: (x−2)²=0. One x-intercept: (2, 0). The vertex sits exactly on the x-axis (Δ=0).
   • (c) y = x² + 2x + 5: Δ = 4 − 20 = −16 < 0. No x-intercepts. The parabola sits entirely above the x-axis.
   • (d) y = 2x² − x − 3: Set y = 0: 2x²−x−3=0. ac=−6, need sum −1: use −3 and 2. 2x²−3x+2x−3=0 ⇒ x(2x−3)+1(2x−3)=0 ⇒ (x+1)(2x−3)=0. x-intercepts: (−1, 0) and (3/2, 0).

3. Full feature analysis of y = x² − 4x + 3. Fluency

   • (a) Direction: a = 1 > 0. Opens upward (minimum turning point).
   • (b) y-intercept: Set x = 0: y = 3. y-intercept: (0, 3).
   • (c) Axis of symmetry: x = −(−4)/(2×1) = 4/2 = 2. Axis: x = 2.
   • (d) Vertex: Substitute x = 2: y = 4−8+3 = −1. Vertex: (2, −1).
   • (e) x-intercepts: Set y = 0: x²−4x+3=0. Factorise: (x−1)(x−3)=0. x-intercepts: (1, 0) and (3, 0).

4. Converting between forms. Fluency

   • (a) y = (x−3)²−4 to standard form: Expand: (x−3)² = x²−6x+9. So y = x²−6x+9−4 = y = x²−6x+5.
   • (b) y = x²−6x+5 to vertex form: Complete the square: y = (x²−6x+9)−9+5 = y = (x−3)²−4. Vertex: (3, −4).
   • (c) y = 2(x+1)²−3 to standard form: Expand: 2(x+1)² = 2(x²+2x+1) = 2x²+4x+2. So y = 2x²+4x+2−3 = y = 2x²+4x−1.
   • (d) y = x²−4x+1 to vertex form: Complete the square: y = (x²−4x+4)−4+1 = y = (x−2)²−3. Vertex: (2, −3).

5. Reading parabolas from equations. Understanding

   • (a) y = 2x²: Opens upward (a = 2 > 0). Vertex at the origin (0, 0). y-intercept (0, 0). x-intercept: x = 0 only (the vertex touches the x-axis). Parabola is narrower than y = x².
   • (b) y = −(x−2)²+4: Opens downward (a = −1). Vertex at (2, 4). y-intercept: x=0 ⇒ y = −4+4 = 0. x-intercepts: (x−2)² = 4 ⇒ x−2 = ±2 ⇒ x = 0 or x = 4. So it crosses at (0,0) and (4,0).
   • (c) y = x²−1: Opens upward. Vertex at (0, −1). y-intercept: (0, −1). x-intercepts: x²=1 ⇒ x = 1 or x = −1. The parabola intersects the x-axis at (−1,0) and (1,0).
   • (d) y = (x+3)²: Opens upward. Vertex at (−3, 0) — the vertex lies on the x-axis. x-intercept: x = −3 only. y-intercept: x=0 ⇒ y = 9 ⇒ (0, 9).

6. Find the equation from features. Understanding

   • (a) Vertex (2,−3) through (0,1): Use vertex form: y = a(x−2)²−3. Substitute (0,1): 1 = a(0−2)²−3 = 4a−3. So 4a = 4 ⇒ a = 1. y = (x−2)²−3 = x²−4x+1.
   • (b) x-ints at x=−1 and x=5, y-int=−5: Use intercept form: y = a(x+1)(x−5). Substitute (0,−5): −5 = a(1)(−5) = −5a ⇒ a = 1. y = (x+1)(x−5) = x²−4x−5.
   • (c) Opens down, vertex (3,7), through (1,3): y = a(x−3)²+7 with a < 0. Substitute (1,3): 3 = a(1−3)²+7 = 4a+7 ⇒ 4a = −4 ⇒ a = −1. y = −(x−3)²+7 = −x²+6x−2.

7. Transformations. Understanding

   • (a) y=x² → y=(x−3)²: Translated 3 units right. Vertex moves from (0,0) to (3,0). Shape and width unchanged.
   • (b) y=x² → y=x²+5: Translated 5 units up. Vertex moves from (0,0) to (0,5). Shape and width unchanged.
   • (c) y=x² → y=−x²: Reflected in the x-axis. The parabola flips from opening upward to opening downward. Vertex stays at (0,0); width unchanged.
   • (d) y=x² → y=3x²: Narrower (vertically stretched by factor 3). Vertex stays at (0,0). For the same x value, y-values are 3 times larger, so the parabola appears steeper.
   • (e) y=x² → y=2(x+1)²−3: Three changes simultaneously: (1) narrower by factor 2; (2) translated 1 unit left; (3) translated 3 units down. Vertex moves from (0,0) to (−1, −3).

8. Ball trajectory. Understanding

   • (a) Initial height: h(0) = 0+0+5 = 5 m.
   • (b) Maximum height: Vertex: t = −4/(2×−1) = −4/−2 = 2 s. h(2) = −4+8+5 = 9 m at t = 2 s.
   • (c) When is h = 8 m?: Set h=8: −t²+4t+5=8 ⇒ −t²+4t−3=0 ⇒ t²−4t+3=0 ⇒ (t−1)(t−3)=0. t = 1 s (going up) and t = 3 s (coming down).
   • (d) When does ball land?: Set h=0: −t²+4t+5=0 ⇒ t²−4t−5=0 ⇒ (t−5)(t+1)=0. t=5 or t=−1. Reject t=−1 (time cannot be negative). Ball lands at t = 5 s.

9. Equation from three points. Problem Solving

   • (a) System of equations: Substituting each point into y=ax²+bx+c: (0,6): c=6. (1,4): a+b+c=4. (3,6): 9a+3b+c=6.
   • (b) Find c: From (0,6): c = 6.
   • (c) Solve for a and b: Substituting c=6 into (1,4): a+b+6=4 ⇒ a+b=−2. Substituting c=6 into (3,6): 9a+3b+6=6 ⇒ 9a+3b=0 ⇒ 3a+b=0. Subtract first from second: (3a+b)−(a+b)=0−(−2) ⇒ 2a=2 ⇒ a=1. Then b=−2−a=−3. a=1, b=−3.
   • (d) Full equation and vertex: y = x²−3x+6. Vertex: x = 3/2 = 1.5. y(1.5) = 2.25−4.5+6 = 3.75. Vertex: (1.5, 3.75).

10. Maximising area. Problem Solving

    • (a) Area expression: Width = x. The two widths use 2x metres, leaving 60−2x for the length. A = x(60−2x) = 60x−2x².
    • (b) Vertex of the parabola: A = −2x²+60x. a = −2 < 0 ⇒ parabola opens downward ⇒ vertex is a maximum. Axis: x = −60/(2×−2) = −60/−4 = x = 15. Vertex: A(15) = −2(225)+900 = −450+900 = 450. Vertex: (15, 450).
    • (c) Optimal dimensions: Width = 15 m. Length = 60−2(15) = 30 m. Maximum area = 450 m².
    • (d) Physically valid range: x must be positive (no negative width) and 60−2x must also be positive (no negative length): 60−2x > 0 ⇒ x < 30. Valid range: 0 < x < 30. At the endpoints (x=0 or x=30), the area drops to zero — no enclosed space.