TMJC H2 Chapter 1 Graphing Techniques Discussion Questions Solutions 2024

Chapter 1 Graphing Techniques TMJC 2024 Page 1 of 20 H2 Mathematics (9758) Chapter 1 Graphing Techniques Discussion Questions Solutions Level 1 1 The curve 1C and 2C have equations e1xy −=+ and ( )ln 2yx=+ respectively. (i) Sketch 1C and 2C on the same diagram, stating the exact coordinates of any points of intersection with the axes and the equations of any asymptotes. (ii) Use your calculator to determine the coordinates of the intersection point(s). 1 Solution (i) e1xy −=+ As x→ , 1y→ Therefore, 1y= is a horizontal asymptote. As x→− , y→ ( )ln 2yx=+ As 2x→− , y→− Therefore, 2x=− is a vertical asymptote. (ii) Using GC, the intersection point is ( )1.44,1.24 to 3 s.f y x O

Chapter 1 Graphing Techniques TMJC 2024 Page 2 of 20 2 Write down the equations of asymptotes of the following graphs: (a) 2 1y x= − (b) 72 3y x=− + (c) 11 2yx x= − − − Hence, sketch the graphs on separate diagrams, indicating equations of asymptotes and coordinates of axial intercepts and turning points (if any). 2 Solution (a) Asymptotes: 1, 0xy== (b) 72 3y x=− + Asymptotes: 3, 2xy=− = y x y = 0 x = 1 𝑦 = 2 1 − 𝑥 (0, 2) y x y = 2 x = −3 ൬0, − 1 3൰ ൬1 2 , 0൰ Recall 1. Asymptotes (if any) 2. Shape of the graph 3. Maximum/minimum turning points (if any) 4. Intercepts with the axes (if any) 5. Labelling of the graph and axes 6. End points (if any)

Chapter 1 Graphing Techniques TMJC 2024 Page 3 of 20 (c) Asymptotes: 2, 1x y x= = − Note For questions that require exact coordinates, students need to find axial intercepts by algebraic method (without GC). i.e. Put 0y= to find x-intercept and put 0x= to find y-intercept. For this question, the x-intercepts are 35 ,02  − and 35 ,02  + . 3 Sketch, on separate diagrams, the following graphs: (a) 22( 3) ( 4) 25xy− + + = (b) 2 2 14 yx += (c) 22 14 16 yx −= (d) ( ) 2 22 14 x y− −= indicating clearly the main relevant features of the graph. Q3 Solution (a) y x x = 2 ( )0,0O ( )0, 8− ( )3, 4− 5 y x ( )6,0 22( 3) ( 4) 25xy− + + = For circle, label - Centre - Radius O ( )0.382,0 10, 2  − Final answer must be 3 s.f. for non-exact values.

Chapter 1 Graphing Techniques TMJC 2024 Page 4 of 20 (b) 2 2 2 2 22114 1 2 y x yx + =  + = (c) 22 22 22 14 16 124 yx yx −=  − = Oblique asymptotes: 22 22 22 22 24 24 1 2 0yx yx yx −= = = 1 2yx= and 1 2yx=− For ellipse, label - Centre - Length of both horizontal/vertical semi-major/semi-minor axis For hyperbola, label - Centre of hyperbola (Intersection of oblique asymptotes) - Coordinates of vertices - Equation of oblique asymptotes To find equation of oblique asymptotes of hyperbola, 1) Change RHS from “1” to “0” (From standard form) 2) Make y the subject 3) Label correctly on diagram by looking at their gradient ( )0, 0 2 1 ( )0, 0

Chapter 1 Graphing Techniques TMJC 2024 Page 5 of 20 (d) To obtain the equations of the asymptotes: ( ) ( ) ( ) 2 2 2 2 2 04 2 4 2 2 11 1 or 1 22 x y xy xy y x y x − −= −= −=  = − =− + 4 Sketch, on separate diagrams, the following graphs: (a) 2,, if 0 2, if 2. xy x x=     (b) 32y x x= − + + 4 Solution (a) y x 𝑦 = − 1 2 𝑥 + 1 𝑦 = 1 2 𝑥 − 1 ሺ𝑥 − 2ሻ2 4 − 𝑦2 = 1 (4, 0) (2, 0) (0, 0) y x

Chapter 1 Graphing Techniques TMJC 2024 Page 6 of 20 (b) −2 O x y 3 5 Exploration for higher order thinking: For ( ) 3 , 33 3 , 3 xxx xx −−= − −  For ( ) 2 , 22 2 , 2 xxx xx + −+= − + −  When 2x− , ( ) ( )2 1 33 2 2 x x xxx− − − + − + + = + =−  When 23 x−   , ( ) ( )33 2 5 2xxxx− − − += +++ =   When 3x , ( ) ( )33 22 21 xx x x x + = + + = −− − + Piecewise graph: 32 1 2 , 2 5 , 2 3 2 1 , 3 y x x xx yx xx = − + + − −  = −    −

Chapter 1 Graphing Techniques TMJC 2024 Page 7 of 20 5 (i) Sketch the curve C defined by the parametric equations 2 1 ,2x y tt== where t is a non-zero real parameter. (ii) Find the Cartesian equation of the curve C. 5 Suggested Solutions (i) 2 1x t= , 2yt= Note: Students are advised to sketch the graph in parametric mode rather than convert parametric to Cartesian before drawing the graph which can be tedious or not possible in some cases. (ii) 2 1x t= , 2yt= 2 yt = 222 1 1 4 2 x ty y = = =  Cartesian equation of C is 2 4xy = O y x Note: When drawing the parametric equations in the GC, go to p and adjust the Tmin and Tmax In this case, since “t is a non-zero real parameter”, you have to include negative values, hence you can set Tmin = −10 and Tmax = 10

Chapter 1 Graphing Techniques TMJC 2024 Page 8 of 20 Level 2 6 H2 Specimen Paper 2006/1/9 Modified Consider the curve 36 ( 6) xy xx −= + . (i) State the coordinates of any points of intersection with the axes. (ii) State the equations of the asymptotes. (iii) Prove, using an algebraic method, that 36 ( 6) x xx − + cannot lie between two certain values (to be determined). (iv) Draw a sketch of the curve, 36 ( 6) xy xx −= + , indicating clearly the main relevant features of the curve. 6 Suggested Solutions (i) (2,0) (ii) Asymptotes: 0, 6xx= =− , 0y= (iii) We first find the region in which the graph can lie i.e. it would have intersection(s) ,y k k= with a horizontal line drawn in that region. Hence, 36 ( 6) xk xx −= + the equation will have real roots . Then we take the complement. 36 ( 6) xy xx −= + Let ,yk= 36 ( 6) xk xx −= + 2 (6 3) 6 0kx k x+ − + = For a quadratic equation to have real roots, discriminant  0 2 2 (6 3) 4 (6) 0 36 60 9 0 kk kk − −  − +  (6 1)(2 3) 0 13 or 62 kk kk − −   Therefore, 36 ( 6) x xx − + cannot lie between 1 6 and 3 2 . + +

Chapter 1 Graphing Techniques TMJC 2024 Page 9 of 20 (iv) Note: - Students need to make connections between parts (i) – (iii) and (iv) to get the correct shape on the right-hand side. - In part (i), there is an intercept at (2,0) so the graph must cut the x-axis at (2,0). - In part (ii), there is a horizontal asymptote y = 0 so the graph must approach the x-axis at the extreme ends. - From (iii), there is no graph between 1 6y= and 3 2y= . So, there should be some graph in the interval 10 6y . In fact, the re is a max turning point with y-coordinate 1 6 . Students may change the window settings to see the shape on the right side: Xmin = 0 Xmax = 20 Ymin = 0 Ymax = 1 x y O y = 0 x = −6 x = 0 (2, 0)

Chapter 1 Graphing Techniques TMJC 2024 Page 10 of 20 7 N2009/1/6 The curve 1C has equation 2 2 xy x −= + . The curve 2C has equation 22 163 xy+= . (i) Sketch 1C and 2C on the same diagram, stating the exact coordinates of any points of intersection with the axes and the equations of any asymptotes. [4] (ii) Show algebraically that the x-coordinates of the points of intersection of 1C and 2C satisfy the equation ( ) ( ) ( ) 22 2 2 2 2 6x x x− = + − . [2] (iii) Use your calculator to find these x-coordinates. [2] 7 Suggested Solutions (i) 1 24:1 22 xCy xx −= = −++ Asymptotes: 1, 2yx= =− 0, 1 0, 2 xy yx = =− == Axial intercepts: (0, –1), (2, 0). 2C is an ellipse with centre (0,0) and x-intercepts ( )6,0 and y-intercepts ( )0, 3 Note: Exact coordinates are required. x y x = −2 y = 1 O (0, −1) (2, 0) Note that C2 passes the point of intersection between the 2 asymptotes of C1.