ACJC 2025 Graphs and Transformations Summary

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Graphs and Transformations / Summary / Page 1 of 5 SUMMARY: Graphs and Transformations 1 Systematic Curve Sketching f( )y x 1. Intercepts Let 0x , find y. Let 0y , find x. 2. Linear Asymptotes If there exists a constant c such that f ( )x  (or  ) as x c , then the line x c is a vertical asymptote of the graph f ( )y x . If there exists a constant k such that f ( )x k as x (and/or x ), then the line y k is a horizontal asymptote of the graph f ( )y x . If g( )y x as x (and/or x ), where g( )x is a linear function, then g( )y x is an oblique asymptote of the graph f ( )y x . 3. Stationary Points  ,f( )a a First Derivative Test x a a a Shape Conclusion d d y x + 0 −  ,f ( )a a is a maximum point. − 0 +  ,f ( )a a is a minimum point. + 0 +  ,f ( )a a is a stationary point of inflexion. − 0 −  ,f ( )a a is a stationary point of inflexion. Second Derivative Test 2 2 d d y x +  ,f ( )a a is a minimum point. −  ,f ( )a a is a maximum point. 0 No conclusion. In this case the firs t derivative test must be done. This does not imply a point of inflexion. 2 Graphs 1. Conic Sections Skill(s) used: Completing the square Circle 2 2 2( ) ( )x h y k r    Ellipse 2 2 2 2 ( ) ( ) 1x h y k a b     Hyperbola 2 2 2 2 ( ) ( ) 1x h y k a b     Parabola     2 y k a x h   2 2 2 2 ( ) ( ) 1y k x h b a         2 x h a y k   x y (h, k) a b a > b O x y (h, k) r O x y (h, k) a b a < b O y x O (h, k) y x O (h, k) (h, k) y x b b O y (h, k) O x a a Note: To label asymptotes!

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Graphs and Transformations / Summary / Page 2 of 5 2 Graphs (cont’d) To obtain asymptotes of hyperbola: 2 2 2 2 ( ) ( ) 1x h y k a b      2 2 2 2 ( ) ( ) 1y k x h b a     As x , 2 2 2 2 ( ) ( )y k x h b a    Thus ( )by k x h a   as x The asymptotes of these hyperbolas are ( )by k x h a   . 2. Modulus Functions f ( ), f ( ) 0f ( ) f ( ), f ( ) 0 x xx x x   1. Keep the positive y portion(s) of f( )y x . 2. Reflect the negative y portion(s) of f( )y x in the x-axis. 3. Delete the negative y portion(s) of f( )y x .   f ( ), 0f f ( ), 0 x xx x x    1. Keep the positive x portion(s) of f( )y x . 2. Delete the negative x portion(s) of f( )y x . 3. Reflect the positive x portion(s) of f( )y x in the y-axis. 3. Rational Functions Skill(s) used: Long division Let P( )f ( ) Q( ) xx x where the degrees of P( )x and Q( )x are less than or equal to 2. If degree of P( )x  degree of Q( )x , then P( )f ( ) Q( ) xx x is a proper fraction.  Horizontal asymptote: x-axis (i.e. 0y ) If degree of P( )x  degree of Q( )x , then P( )f ( ) Q( ) xx x is an improper fraction.  Long division  R( )f ( ) Q( ) xx px q x   , where R( ) Q( ) x x is a proper fraction  0p  horizontal asymptote y q ; 0p  oblique asymptote y px q  In both cases, solve Q( ) 0x  to find vertical asymptote(s) ax by cx d   asymptotes: dx c , ay c   2 , 0ax bx c ry px q adx e dx e        asymptotes: ex d  , y px q  f ( )y x f ( )y x  fy x

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Graphs and Transformations / Summary / Page 3 of 5 3 Simple Transformations The following table summarises simple transformations for the case where 0a . Transformation Description of transformation Geometrically TRANSLATION f ( ) f ( )y x y x a    Replace x by x – a. The graph of f ( )y x a  is the translation of the graph of f ( )y x by a units in the positive x-direction.    , ,x y x a y  f ( ) f ( )y x y x a    Replace x by x + a. The graph of f ( )y x a  is the translation of the graph of f ( )y x by a units in the negative x-direction.    , ,x y x a y  f ( ) f ( ) i.e. f ( ) y x y a x y x a       Replace y by y – a. The graph of f ( )y x a  is the translation of the graph of f ( )y x by a units in the positive y-direction.    , ,x y x y a  f ( ) f ( ) i.e. f ( ) y x y a x y x a       Replace y by y + a. The graph of f ( )y x a  is the translation of the graph of f ( )y x by a units in the negative y-direction.    , ,x y x y a  SCALING f ( ) f xy x y a        Replace x by x a . The graph of f xy a      is the scaling of the graph of f ( )y x by factor a parallel to the x-axis.    , ,x y ax y f ( ) f ( ) i.e. f ( ) yy x x a y a x     Replace y by y a . The graph of f ( )y a x is the scaling of the graph of f ( )y x by factor a parallel to the y-axis.    , ,x y x ay REFLECTION f ( ) f ( )y x y x    Replace x by – x. The graph of f ( )y x  is the reflection of the graph of f ( )y x in the y-axis.    , ,x y x y  f ( ) f ( )y x y x   Replace y by – y. The graph of f ( )y x is the reflection of the graph of f ( )y x in the x-axis.    , ,x y x y 

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Graphs and Transformations / Summary / Page 4 of 5 4 Graph of 1 f( )y x Graph of f ( )y x Graph of 1 f( )y x Let b  0: f ( )y x has an x-intercept at x a . 1 f ( )y x has a vertical asymptote x a . f ( )y x has a vertical asymptote x a . 1 f ( )y x has an x-intercept x a . f ( )y x has a y-intercept at y b . 1 f ( )y x has a y-intercept at 1y b . f ( )y x has a horizontal asymptote y b . 1 f ( )y x has a horizontal asymptote 1y b . (a) As f ( )x b  , (a) 1 1 f ( )x b      . (b) As f ( )x b  , (b) 1 1 f ( )x b      . f ( )x  . 1 0f ( )x  . f ( )x  . 1 0f ( )x  . ( , )a b is a maximum point. 1,a b      is a minimum point. ( , )a b is a minimum point. 1,a b      is a maximum point. For checking: f ( )x increases. 1 f ( )x decreases. f ( )x decreases. 1 f ( )x increases. f ( ) 0x  . 1 0f ( )x  . f ( ) 0x  . 1 0f ( )x  . f ( )y x has an oblique asymptote. 1 f ( )y x has a horizontal asymptote 0y .

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Graphs and Transformations / Summary / Page 5 of 5 4 Graph of 1 f( )y x Graph of f( )y x Graph of 1 f( )y x 5 Graphs of Parametric Equations  f( ), f( )x t y t 1. Work with parametric equations Key in the pair of parametric equations into GC and sketch directly via PAR mode. 1. Change calculator to PAR mode. 2. Ensure RADIANS mode. 3. Key in expressions for x and y. 4. Change window settings for range of t. Default window settings are: tmin  0, tmax  2π. 2. Convert to Cartesian equation Convert parametric equations into Cartesian equation by eliminating the parameter t and use GC to sketch via FUNC mode. Type I: Make t the subject using one parametric equation, then do substitution into the other parametric equation Type II: Algebraic manipulations, e.g. squaring each parametric equation then adding / subtracting the two equations; adding and subtracting both parametric equations then eliminate t; etc. Type III: Use trigonometric identities (a) 2 2sin cos 1   (b) 2 2tan 1 sec   (c) 2 2cot 1 cosec   and others in MF27 3. Find the point(s) of intersection of a curve (parametric) and a line (Cartesian) Substitute the pair of parametric equations into the Cartesian equation to solve for the parameter t. Then obtain the values of x and y from the parametric equations using the value(s) of t found. Curve: 3 2x t  and tty 3 ----- (1) Line: 9 2 14y x  ----- (2) Sub (1) into (2),  39 2 3 2 14t t t        027183 2  tt   033 2 