O-Level Quadratic functions

y = 4x - 4x² - 3 = -4x² + 4x - 3 = -4(x² - x) - 3 = -4[(x - 1/2)² - 1/4] - 3 = -4(x - 1/2)² + 1 - 3 = -4(x - 1/2)² - 2 So, comparing with p(x + q)² + r: p = -4, q = -1/2, r = -2. Check: at x = 0, -4(1/4) - 2 = -3, and 4(0) - 4(0) - 3 = -3. Correct.

"Can never be less than 1" means the smallest value the expression ever takes is still 1 or more. Completing the square is the method that reveals that smallest value, because it rewrites the expression as a square (which cannot be negative) plus a constant. Step 1: take the coefficient of x² out of the x terms only, leaving the constant alone for now. 3x² + 15x + 20 = 3(x² + 5x) + 20 Step 2: complete the square inside the bracket. Halve the coefficient of x: half of 5 is 2.5. Then x² + 5x = (x + 2.5)² − 2.5² and 2.5² = 6.25, so x² + 5x = (x + 2.5)² − 6.25 Substituting this back: 3x² + 15x + 20 = 3[(x + 2.5)² − 6.25] + 20 Step 3: expand the 3 across the bracket and collect the constants. = 3(x + 2.5)² − 18.75 + 20 = 3(x + 2.5)² + 1.25 Step 4: now reason about the size of each part. (x + 2.5)² is a square of a real number, so (x + 2.5)² ≥ 0 for every value of x. Multiplying by 3, which is positive, keeps the direction of the inequality: 3(x + 2.5)² ≥ 0. Adding 1.25 to both sides: 3(x + 2.5)² + 1.25 ≥ 1.25. Step 5: so the expression is always at least 1.25, and the least value 1.25 occurs when x = −2.5. Since 1.25 > 1, the expression 3x² + 15x + 20 can never be less than 1.

'Greater than zero for every x' means the whole parabola lies above the x-axis, which needs two conditions to hold together. Condition 1: the curve opens upwards, so the coefficient of x² must be positive: m − 6 > 0 m > 6 Condition 2: the curve never touches or crosses the x-axis, so (m − 6)x² − 8x + m = 0 has no real roots, which means the discriminant is negative. Here a = m − 6, b = −8 and c = m: (−8)² − 4(m − 6)(m) < 0 64 − 4m(m − 6) < 0 64 − 4m² + 24m < 0 Divide every term by −4, and REVERSE the inequality sign because it is a division by a negative: −16 + m² − 6m > 0 m² − 6m − 16 > 0 Factorise: two numbers multiplying to −16 and adding to −6 are −8 and +2: (m − 8)(m + 2) > 0 The critical values are m = 8 and m = −2, and a U-shaped expression is positive OUTSIDE its roots: m < −2 or m > 8 Combine the two conditions. m > 6 AND (m < −2 or m > 8). Nothing can be both greater than 6 and less than −2, so that branch is rejected. m > 8