3.6 Quadratic Functions, SPM Practice (Short Questions)


3.6 Quadratic Functions, SPM Practice (Short Questions)

Question 1:
Find the minimum value of the function f (x) = 2x2 + 6x + 5. State the value of x that makes f (x) a minimum value.

Solution:
By completing the square for f (x) in the form of f (x) = a(x + p)2 + q to find the minimum value of f (x).

f( x )=2 x 2 +6x+5 =2[ x 2 +3x+ 5 2 ] =2[ x 2 +3x+ ( 3× 1 2 ) 2 ( 3× 1 2 ) 2 + 5 2 ]  
=2[ ( x+ 3 2 ) 2 9 4 + 5 2 ] =2[ ( x+ 3 2 ) 2 + 1 4 ] =2 ( x+ 3 2 ) 2 + 1 2  

Since a = 2 > 0,
Therefore f (x) has a minimum value when x= 3 2 .  The minimum value of f (x) = ½


Question 2:
The quadratic function f (x) = –x2 + 4x + k2, where k is a constant, has a maximum value of 8.
Find the possible values of k.

Solution:
f (x) = –x2 + 4x + k2
f (x) = –(x2 – 4x) + k2 ← [completing the square for f (x) in
the form of f (x) = a(x + p)2 + q]
f (x) = –[x2 – 4x + (–2)2 – (–2)2] + k2
f (x) = –[(x – 2)2 – 4] + k2
f (x) = –(x – 2)2 + 4 + k2

Given the maximum value is 8.
Therefore, 4 + k2 = 8
k2 = 4
k = ±2