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2.12.6 Quadratic Functions, SPM Practice (Paper 1)


Question 17:
Find the minimum value of the function f (x) = 2x2 + 6x + 5. State the value of xthat 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 + 6 x + 5 = 2 [ x 2 + 3 x + 5 2 ] = 2 [ x 2 + 3 x + ( 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 18:
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


Question 19:
Find the maximum value of the function 5 – x – 2x2 , and the corresponding value of x.

Solution:
5x2 x 2 =2 x 2 x+5 =2[ x 2 + 1 2 x 5 2 ] =2[ x 2 + 1 2 x+ ( 1 4 ) 2 ( 1 4 ) 2 5 2 ] =2[ ( x+ 1 4 ) 2 1 16 5 2 ] =2[ ( x+ 1 4 ) 2 41 16 ] =2 ( x+ 1 4 ) 2 +5 1 8


5x2 x 2  has a maximum value when 2 ( x+ 1 4 ) 2 =0   x= 1 4 The maximum value of 5x2 x 2  is 5 1 8 .

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