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 to response to the question. Learning point 1: To find vertical asymptote, let denominator 0x b x b− =  = Learning point 2: To determine the values a rational function can take, introduce k to be these values and ensure the quadratic equation formed has real roots Please answer the question directly.

Chapter 1 Graphing Techniques TMJC 2024 Page 4 of 9 (iii) (iv) 22 22 10( 1) x axxc xb  ++− + = − 2 2 2( 1)x y c − + = Add the graph 2 2 2( 1)x y c−+= into the previous part, From the graph, 3c Explanation We are finding the range of values of c such that there is at least one point of intersection between 2 5 10 3 xxy x −+= − and a circle with centre ( )1,0 and radius c (since 0c from the question). (5, 5) y x O (1, −3) ൬0, − 10 3 ൰ x = 3 y = x − 2 𝑦 = 𝑥2 − 5𝑥 + 10 𝑥 − 3 03 c When , no point of intersection between 2 5 10 3 xxy x −+= − and ( ) 2 221x y c− + = Learning point 3: For 22 22 10( 1) x axxc xb  ++− + = − to have at least one real root is equivalent to having at least one point of intersection between 2 5 10 3 xxy x −+= − and ( ) 2 221x y c− + = .

Chapter 1 Graphing Techniques TMJC 2024 Page 5 of 9 4 2020/RVHS/JC2 MYE/Q4a(i) The curve C has parametric equations 1 2sinxt=+ , cosyt= , for ππ .22 t−   (i) Sketch C, indicating clearly the exact coordinates of the axial intercepts. [2] (ii) Find the cartesian equation of C. [1] 4 Suggested Solutions 3c= When , one point of intersection between 2 5 10 3 xxy x −+= − and ( ) 2 221x y c− + = 3c When , more than one point of intersection between 2 5 10 3 xxy x −+= − and ( ) 2 221x y c− + = Change the window settings according to the range of values of t given in question: ππ 22 t−  

Chapter 1 Graphing Techniques TMJC 2024 Page 6 of 9 Solving for y-intercept: When 0,x= 1 2sin 0t+= 1sin 26t  =− − 3cos 62y  = − = 11 2sin sin --- (1) 2 xx t t −= +  = cos --- (2)yt= Substitute (1) and (2) into 22sin cos 1tt+= , ( ) 2 2 2 2 1 12 1 14 x y x y − += − += 1 1 1sin 12 6 1sin 26   − − − =−  − =− 2 3cos 64 33cos 6 4 2    −=  − = = Note: (1) Sketch the curve according to the range of values of t given in question. (2) Label end-points which are the x- intercepts in this case (3) Label in coordinates as indicated in question (4) Indicate open/closed circle for end points according to range of values of t. In this case, both end-points are closed circle because ππ 22 t−   . A good habit to label the corresponding values of t on the graph itself for easy reference in subsequent parts of the question. Generally, if the parametric equation has trigonometry, you will be using trigo identities like sin2 𝜃 + cos2 𝜃 = 1 to convert to cartesian equation Note: Only convert parametric equations to cartesian equation if the question asks for it

Chapter 1 Graphing Techniques TMJC 2024 Page 7 of 9 5 The parametric equations of a curve are 2 2,x t y t== for , 0.tt Sketch C. [2] 5 Suggested Solutions 6 2010/A-Level/P1/Q11(iii) – Challenge yourself A curve C has parametric equations 11,.x t y t tt= + = − Find a cartesian equation of C. Sketch C, giving the coordinates of any points where C crosses the x- and y-axes and the equations of any asymptotes. [4] 6 Suggested Solutions ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 22 22 22 1 1 1 2 1 2 : 2 3 21 2 : 4 3 4 : 4 The cartesian equation is 4 122 xt t yt t x y t xy t x y x y xy xy = + −−−−−−−−− = − −−−−−−−−− + + = −−−−−−−−− − − = −−−−−−−−−  + − = −= −= O y x Since 𝑡 ∈ ℝ, 𝑡 ≠ 0, it means t can be any real number except zero Change the window settings to include negative values to observe the complete graph. T min 10 T max 10 =− = Observing the asymptotes: As , , 0x t y→ → → Thus, 0y= is an asymptote. As 0, 0,x t y→ → → Thus, 0x= is an asymptote. The idea to obtain the cartesian equation from the parametric equations is to eliminate the parameter t.

Chapter 1 Graphing Techniques TMJC 2024 Page 8 of 9 22 22 22 To find the equation of the oblique asymptotes, let 022 xy yx yx −= = = 2 2 When 0, 12 2 y x x = = = Do not forget the “  ” y x O

Chapter 1 Graphing Techniques TMJC 2024 Page 9 of 9 7 Match the following equations with their corresponding sketches, and fill in the boxes accordingly. 1. 2 16 4y x=+ − 6. 22 2( ) 0x y x y+ − − = 2. 22 136 9 yx−= 7. 531 4yx x= − + − 3. 25 ( 2)yx− = + 8. 22 125 16 xy−= 4. 22( 3) ( 4) 25xy− + − = 9. 224 8 2 4x y x y+ − − = 5. 2 2 14 x y+= 10. 2( 4) 2yx− = − G D F E J B