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2.13.2 Quadratic Functions, SPM Practice (Paper 2 Question 6 – 10)


Question 7:
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 8:

The diagram above shows the graph of a quadratic function y = f(x). The straight line y = –4 is  tangent 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 or x ≥ 6 ← (above x-axis).



Question 9:
Given that the quadratic function f(x) = 2x2 – px + p has a minimum value of –18 at x = 1.
(a) Find the values of p and q.
(b) With the value of p and q found in (a), find the values of x, where graph f(x) cuts the x-axis.
(c) Hence, sketch the graph of f(x).

Solution:
(a)
f( x )=2 x 2 −px+q =2[ x 2 − p 2 x+ q 2 ] =2[ ( x+ −p 4 ) 2 − ( −p 4 ) 2 + q 2 ] =2[ ( x− p 4 ) 2 − p 2 16 + q 2 ] =2 ( x− p 4 ) 2 − p 2 8 +q


∴ p 4 =1−−−−−−( 1 ) and − p 2 8 +q=−18−−−( 2 ) From( 1 ),p=4. Substitute p=4 into ( 2 ): − ( 4 ) 2 8 +q=−18   − 16 8 +q=−18  q=−18+2    =−16


(b)
f( x )=2 x 2 −4x−16 f( x )=0 when it cuts x-axis 2 x 2 −4x−16=0 x 2 −2x−8=0 ( x−4 )( x+2 )=0 x=4,−2 Graph f( x ) cuts x-axis at x=−2 and x=4.

(c)


Question 10:
(a) Find the range of values of k if the equation x2 – kx + 3k – 5 = 0 does not have real roots.
(b) Show that the quadratic equation hx2 – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Solution:
(a)
x 2 −kx+( 3k−5 )=0 If the above equation has no real root, ∴  b 2 −4ac<0. k 2 −4( 3k−5 )<0 k 2 −12k+20<0 ( k−2 )( k−10 )<0

Graph function y = (k – 2)(k – 10) cuts the horizontal line at k = 2 and k = 10 when b2 – 4ac < 0.



The range of values of k that satisfy the inequality above is 2 < k < 10.


(b)
h x 2 −( h+3 )x+1=0 b 2 −4ac= ( h+3 ) 2 −4( h )( 1 ) = h 2 +6h+9−4h = h 2 +2h+9 = ( h+ 2 2 ) 2 − ( 2 2 ) 2 +9 = ( h+1 ) 2 −1+9 = ( h+1 ) 2 +8

The minimum value of (h + 1) + 8 is 8, a positive value. Therefore, b2 – 4ac > 0 for all values of h.
Hence, quadratic equation hx2 – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

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