RI C8B Curve Transformation Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________ Tutorial 8B : Curve Transformation Page 1 of 4 Tutorial 8B: Curve Transformation Section A (Basic Questions) 1 The diagram above shows the graph of f( )yx . The curve passes through the points (0, 1)A , (2, 2)B and (3,0)C . Sketch on separate diagrams, the graphs of (i) f( 1 )yx , (ii) f( ) 1yx , (iii) f( 2 )yx , (iv) f 2xy   (v) 3f ( )yx , (vi) f( )yx , (vii) f( )yx , showing in each case, the coordi nates of points corresponding to A, B and C. Solution: (i) f( 1 )yx (ii) f( ) 1yx x y O y x O 1(3, 2)B 1 (4,0)C 1(1,1)A y x O 2 (2,1)B 2 (0,0)A 2 (3, 1)C 

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _____________________________ Tutorial 8B: Curve Transformation Page 2 of 4 (iii) f( 2 )yx (iv) f 2xy   (v) 3f ( )yx (vi)  fyx (vii)  fyx y x O 3 (0,1)A 3 3 ,02C  3 (1, 2)B y x O x y O y x O 5 (0, 3)A  5 (2, 6)B  5 (3,0)C 5 (0,1)A 4 (4,2)B 5 (6,0)B x y

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _____________________________ Tutorial 8B: Curve Transformation Page 3 of 4 x y 2 2 O 2 For each of the following graphs with equation y = f(x), sketch, on separate diagrams, the graphs of (i) fyx  , (ii)  1 fy x (a) The function f is a semi-circle with domain being [ 2, 2]. (0, 2) is a maximum turning point. (b) Domain of f is ( , ). Maximum gradient occurs at x = 0 and 2y are horizontal asymptotes. (c) Domain of f is 1\ 2 . 11 and are asymptotes.22yx  x y O 2 y x O 1

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _____________________________ Tutorial 8B: Curve Transformation Page 4 of 4 Solution: (a)  fyx   1 fy x (b)  1 fy x (c) x y O y = f (x) x y O Max gradient y = f (x) 0.5 0.5y 0 x y 0 x y 1 0.5 1 1x  1 fy x