Finding the polynomial: Let y = ax² + bx + c Point (0, -4): c = -4 ... (equation 1) Point (1, -3): a + b + c = -3 Substitute c = -4: a + b - 4 = -3 → a + b = 1 ... (equation 2) Point (-1, -3): a - b + c = -3 Substitute c = -4: a - b - 4 = -3 → a - b = 1 ... (equation 3) Adding equations (2) and (3): 2a = 2 → a = 1 From equation (2): b = 1 - a = 1 - 1 = 0 Part (a): Polynomial is y = x² - 4 Part (b): Direction of graph Coefficient of x² is a = 1 > 0 Graph opens upward Part (c): Finding zeroes x² - 4 = 0 x² = 4 x = ±2 Zeroes: x = 2 and x = -2For polynomial x² - 2x - 8: Sum of zeroes: α + β = -(-2)/1 = 2 Product of zeroes: αβ = -8/1 = -8 Finding α²/β² + β²/α²: α²/β² + β²/α² = (α⁴ + β⁴)/(αβ)² Step 1: Find α² + β² α² + β² = (α + β)² - 2αβ = (2)² - 2(-8) = 4 + 16 = 20 Step 2: Find α⁴ + β⁴ α⁴ + β⁴ = (α²)² + (β²)² = (α² + β²)² - 2(αβ)² = (20)² - 2(-8)² = 400 - 2(64) = 400 - 128 = 272 Step 3: Calculate final value α²/β² + β²/α² = (α⁴ + β⁴)/(αβ)² = 272/(-8)² = 272/64 = 17/4 Answer: 17/4 or 4.25Part (a): Finding the polynomial Let f(x) = x² + bx + c (coefficient of x² = 1) Given: One zero is 3, so (x - 3) is a factor Let other zero be α f(x) = (x - 3)(x - α) = x² - (3 + α)x + 3α So b = -(3 + α), c = 3α Given: f(1) = -6 1 + b + c = -6 → b + c = -7 Substitute: -(3 + α) + 3α = -7 -3 - α + 3α = -7 2α = -4 → α = -2 Therefore: b = -(3 + (-2)) = -1, c = 3(-2) = -6 Polynomial: f(x) = x² - x - 6 Part (b): Finding other zero From above: α = -2 Or by factoring: x² - x - 6 = (x - 3)(x + 2) = 0 Other zero: -2 Part (c): Discriminant and graph analysis Discriminant D = b² - 4ac = (-1)² - 4(1)(-6) = 1 + 24 = 25 Since D = 25 > 0 Number of zeroes = 2 (graph intersects x-axis at two points)For polynomial y = kx² - 4x + k Compare with ax² + bx + c: a = k, b = -4, c = k Discriminant D = b² - 4ac = (-4)² - 4(k)(k) = 16 - 4k² Part (a): Condition for two distinct points (two distinct zeroes) Condition: D > 0 16 - 4k² > 0 4k² < 16 k² < 4 -2 < k < 2 But k ≠ 0 (otherwise not a quadratic) Answer: -2 < k < 2 and k ≠ 0 Part (b): Touching at exactly one point (equal roots) Condition: D = 0 16 - 4k² = 0 4k² = 16 k² = 4 k = ±2 Answer: k = 2 or k = -2 Part (c): No real zeroes (no x-intercepts) Condition: D < 0 16 - 4k² < 0 4k² > 16 k² > 4 k < -2 or k > 2 Answer: k < -2 or k > 2Let the three zeroes be: α, -α, and β (Two zeroes equal in magnitude but opposite in sign) Using sum of zeroes: For polynomial x³ - 5x² - 16x + 80 Compare with Ax³ + Bx² + Cx + D: A = 1, B = -5, C = -16, D = 80 Sum of zeroes = -B/A = -(-5)/1 = 5 α + (-α) + β = 5 0 + β = 5 β = 5 Using product of zeroes: Product of zeroes = -D/A = -80/1 = -80 α × (-α) × β = -80 -α² × β = -80 -α² × 5 = -80 α² = 16 α = ±4 All zeroes: 4, -4, 5
