3.9.4 Quadratic Functions, SPM Practice (Long Question)


Question 6:
Quadratic function f(x) = x2 – 4px + 5p2 + 1 has a minimum value of m2 + 2p, where m and p are constants.
(a) By using the method of completing the square, shows that m = p – 1.
(b) Hence, find the values of p and of m if the graph of the quadratic function is symmetry at x = m2 – 1.

Solution:
(a)
f( x )= x 2 4px+5 p 2 +1 = x 2 4px+ ( 4p 2 ) 2 ( 4p 2 ) 2 +5 p 2 +1 = ( x2p ) 2 + p 2 +1 Minimum value, m 2 +2p= p 2 +1 m 2 = p 2 2p+1 m 2 = ( p1 ) 2 m=p1

(b)
x= m 2 1 2p= m 2 1 p= m 2 1 2 Given m=p1p=m+1 m+1= m 2 1 2 2m+2= m 2 1 m 2 2m3=0 ( m3 )( m+1 )=0 m=3 or 1 When m=3, p= 3 2 1 2 =4 When m=1, p= ( 1 ) 2 1 2 =0