3.7 Quadratic Functions, SPM Practice (Long Questions)


3.7 Quadratic Functions, SPM Practice (Long Questions)
Question 1:
Without drawing graph or using method of differentiation, find the maximum or minimum value of the function y = 2 + 4x – 3x2. Hence, find the equation of the axis of symmetry of the graph.

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

y = 2 + 4x – 3x2
y = – 3x2 + 4x + 2 ← (in general form)
y=3[ x 2 4 3 x 2 3 ] y=3[ x 2 4 3 x+ ( 4 3 × 1 2 ) 2 ( 4 3 × 1 2 ) 2 2 3 ] y=3[ ( x 2 3 ) 2 ( 2 3 ) 2 2 3 ]  

y=3[ ( x 2 3 ) 2 4 9 6 9 ] y=3[ ( x 2 3 ) 2 10 9 ] y=3 ( x 2 3 ) 2 + 10 3 in the form of a (x+p) 2 +q

Since a = –3 < 0,
Therefore, the function y has a maximum value of 10 3 .  
x 2 3 =0 x= 2 3
Equation of the axis of symmetry of the graph is x= 2 3 .   



Question 2:

The diagram above shows the graph of a quadratic function y = f(x). The straight line y = –4 is  tangen to the curve y = f(x).
(a)    Write the equation of the axis of symmetry of the function f(x).
(b)   Express f(x) in the form of (x + p)2 + q , where p and q are constant.
(c)    Find the range of values of x so that
(i)     f(x) < 0,            (ii) f(x) ≥ 0.

Solution:
(a)
x-coordinate of the minimum point is the midpoint of (–2, 0) and (6, 0)
= 2+6 2 =2  
Therefore, equation of the axis of symmetry of the function f(x) is x = 2.

(b)
Substitute x = 2 into x + p = 0,
2 + p = 0
p = –2
and q = –4 (the smallest value of f(x))
Therefore, f(x) = (x + p)2 + q
     f(x) = (x – 2)2 – 4

(c)(i) From the graph, for f(x) < 0, range of values of x are –2 < x < 6 ← (below x-axis).

(c)(ii) From the graph, for f(x) ≥ 0, range of values of x are x ≤ –2 atau x ≥ 6 ← (above x-axis).