3.6 Quadratic Functions, SPM Practice (Long Question)


Question 3:
Given that the quadratic function f(x) = 2x2px + 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 2 +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 4x16 f( x )=0 when it cuts x-axis 2 x 2 4x16=0 x 2 2x8=0 ( x4 )( x+2 )=0 x=4,2 Graph f( x ) cuts x-axis at x=2 and x=4.

(c)



Question 4:
(a) Find the range of values of k if the equation x2kx + 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+( 3k5 )=0 If the above equation has no real root,   b 2 4ac<0. k 2 4( 3k5 )<0 k 2 12k+20<0 ( k2 )( k10 )<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+94h = 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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