O-Level Quadratic functions: worked solution

Question

The curve y=px²−(p+2)x+1 has constant p>0. (a) Complete the square to obtain the least value of y, written in the form (a−p²)/(bp).

Complete the square on y = px² − (p + 2)x + 1. The coefficient of x² is p, not 1, so first take p out as a factor of the two terms that contain x: y = p[x² − ((p + 2)/p)x] + 1. Inside the bracket, halve the coefficient of x and square it. Half of (p + 2)/p is (p + 2)/(2p), so x² − ((p + 2)/p)x = (x − (p + 2)/(2p))² − ((p + 2)/(2p))². The subtracted square is ((p + 2)/(2p))² = (p + 2)²/(4p²). Multiply the whole bracket back by p: y = p(x − (p + 2)/(2p))² − p × (p + 2)²/(4p²) + 1 y = p(x − (p + 2)/(2p))² − (p + 2)²/(4p) + 1. Because p > 0, the term p(x − (p + 2)/(2p))² is never negative, and it is exactly zero when x = (p + 2)/(2p). So the least value of y is whatever is left when that term vanishes: least y = 1 − (p + 2)²/(4p). Write this as a single fraction over the common denominator 4p: = [4p − (p + 2)²] / (4p). Expand (p + 2)² = p² + 4p + 4: = [4p − (p² + 4p + 4)] / (4p) = (4p − p² − 4p − 4) / (4p) = (−p² − 4) / (4p) = (−4 − p²) / (4p). Comparing with the required form (a − p²)/(bp) gives a = −4 and b = 4.