TMJC H2 Chapter 1 Graphing Techniques Assignment_Solutions

Chapter 1 Graphing Techniques TMJC 2024 Page 1 of 9 H2 Mathematics (9758) Chapter 1 Graphing Techniques Assignment Solutions 1 2017/NYJC Promo/4 The curve C has equation 242 21 xy x −= + . Sketch C, giving the exact coordinates of all points of intersection with the axes and the equations of the asymptotes. [3] Q1 Solution 24 2 1 212 1 2 1 xyx xx −= = − −++ y x y = 2x − 1 𝑥 = − 1 2 ቆ− ξ2 2 , 0ቇ ቆξ2 2 , 0ቇ (0, −2) Please read the requirements of question carefully. Question stated “exact coordinates”, hence you should not write the x- ( )0.707,0 intercepts as and ( )0.707,0− . Always make the rational function into proper fraction using long division or juggling first, to correctly identify the asymptotes.

Chapter 1 Graphing Techniques TMJC 2024 Page 2 of 9 2 Sketch the graph of 22 4 2 8 1 0x y x y+ + − + = . State the line(s) of symmetry. Q2 Solution ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2 22 22 2 22 2 2 22 22 4 2 8 1 0 1 1 4 2 1 0 1 4 1 1 0 1 4 1 4 1 114 11 121 x y x y x y y xy xy x y xy + + − + = + − + − + = + + − − = + + − = + + − = +− + = The lines of symmetry are 1x=− and 1y= . Note: It is an ellipse, centre at ( )1,1− with horizontal semi-major axis of length 2 units and vertical semi-minor axis of length 1 unit. y x (−1,0) (−1,1) 1 2 Read the question! Question asks to state the lines of symmetry, so you have to state it clearly Note: you do not need to label the intercepts if the centre is not on the axis. However, in this case, the x-intercept can be easily identified hence you should label it For ellipse, you will need to label the centre and radius clearly.

Chapter 1 Graphing Techniques TMJC 2024 Page 3 of 9 3 2018/MJC Promo/3 (Modified) The curve C has equation 2 10x axy xb ++= − . The asymptotes of C are 3x= and 2.yx=− (i) State the value of b and show that 5.a=− [2] (ii) Using an algebraic method, determine the values 2 10x ax xb ++ − can take. [3] (iii) Sketch C, showing its asymptotes and the coordinates of axial intercept and turning points. [3] (iv) Deduce the range of values of c, where c is a positive real constant such that the equation 22 22 10( 1) x axxc xb  ++− + = − has at least one real root. [2] Q3 Solution (i) b = 3 ( )( ) 2 2 22 102 3 2 3 10 5 6 10 d x axx x x b x x d x ax x x d x ax ++− + =−− − − + = + + − + + = + + By comparing coefficient, 5a=− (ii) 2 5 10 3 xxy x −+= − . Let 2 5 10,. 3 xxy k k x −+== − ( ) 2 5 10 3 0x k x k− + + + = For quadratic equation to have real roots, discriminant 0 . ( )( ) ( )( ) ( )( ) 2 2 2 5 4 1 10 3 0 10 25 40 12 0 2 15 0 5 3 0 kk k k k kk kk − + − +  + + − −  − −  − +  3 or 5kk−  Therefore, 2 5 10 3 xx x −+ − can take values less than or equal to 3− or values more than or equal to 5. You can still use the condition discriminant < 0, but you will need to take the complement of the answer