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)=x2−4px+5p2+1=x2−4px+(−4p2)2−(−4p2)2+5p2+1=(x−2p)2+p2+1Minimum value,m2+2p=p2+1m2=p2−2p+1m2=(p−1)2m=p−1
(b)
x=m2−12p=m2−1p=m2−12Given m=p−1⇒p=m+1m+1=m2−122m+2=m2−1m2−2m−3=0(m−3)(m+1)=0m=3 or −1When m=3,p=32−12=4When m=−1,p=(−1)2−12=0
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)=x2−4px+5p2+1=x2−4px+(−4p2)2−(−4p2)2+5p2+1=(x−2p)2+p2+1Minimum value,m2+2p=p2+1m2=p2−2p+1m2=(p−1)2m=p−1
(b)
x=m2−12p=m2−1p=m2−12Given m=p−1⇒p=m+1m+1=m2−122m+2=m2−1m2−2m−3=0(m−3)(m+1)=0m=3 or −1When m=3,p=32−12=4When m=−1,p=(−1)2−12=0