Given graph of y f = and sketch:. Linear Transformation cf ( b + a) + d a. translate a along the -ais. f b. scale b along the -ais c. scale c along the y-ais d. translate d along the y-ais Transformation of curve a. reflect the portion of the curve that is below the -ais about the -ais 3. f ( ) a. keep only portion of the curve y f = that is in the 0 region b. reflect that portion about the y-ais 4. f ( ) a. keep only portion of the curve y f = that is in the 0 region b. reflect that portion about the y-ais 5. y = f a. Sketch f i. keep portion of y f Optional = that is above the -ais. Square root all y values ii. Draw iii. The new curve is above the old curve for region 0 < y < iv. The new curve is below the old curve for region y > v. The new and old curve will intersect at 0 and b. Reflect f about the -ais to get f c. If the curve has a turning point at the -ais, it will be a sharp edge else it is flat. stationar y point non stationary point sharp flat Page of 8
6. f a. all y values. y b. Draw the line i. The new curve is above old curve for 0 < y <. ii. The new curve is below the old curve y > c. Draw the line i. The new curve is above below curve for < y < 0. ii. The new curve is above the old curve y < d. The vertical asymptotes become -ais intercepts e. The -ais intercepts become vertical asymptotes f. The maimum point becomes minimum point g. The minimum point become maimum points h. Analysis small and large values. Optional 7. f ' a. Vertical asymptotes = a remains b. Horizontal asymptotes k becomes k = 0 c. Oblique asymptotes a + b becomes a d. Stationary points becomes -intercepts e. Stationary points of inflection become turning points. y f f. Find a few points on =, estimate the gradient and plot out on the y f ' g. Connect the points. Note: Highlighted portion are optional. Advance concepts : Composite transformations For some questions, you may be required to do a series of transformations to get the final f + answers. Eample: a) Sketch f ( + ) = g b) Sketch g ( ) ( g ( ) f ( ) = + ) =. Page of 8
Finding equations/ describing transformations Let original equation of a curve be y f equation of the curve be g. Type : The final curve y g =, after a series of transformation, the final = is unknown, find it. Apply the transformations one at a time replacing the variables correctly each time. Eample 00/I/5 (part of) 3 The curve with equation is transformed by a translation of units in the positive - direction, followed by a stretch with scale factor parallel to the y-ais, followed by a translation of 6 units in the negative y-direction. Find the equation of the new curve in the g form Let 3 = f After translation of units in the positive -direction: Replace with ( ) 3 After a stretch with scale factor parallel to the y-ais: Replace with 3 3 ( ) ( ) After a translation of 6 units in the negative y-direction: Replace with +6 3 y + 6 = ( ) 3 6 Hence ( ) 3 6 is the final equation of graph. Page 3 of 8
Type : Given f and y g = describe the transformations. Find a possible transformation a step at a time. At every transformation, replace the variables correctly each time. Note: There is no standard answer for this type of questions. Eample 007/I/5 part of + 7 B Show that the equation can be written as A +, where and are + + constants to be found. Hence state a sequence of transformations which transform the graph + 7 of to the graph of. + + 7 3 = + + + Original curve = f A: Translation of units in the negative direction of -ais. Replace with + + B: Scaling with a factor of 3 parallel to the y-ais. Replace with y 3 = 3 + + C: Translation of units in the direction of positive y-ais. Replace with 3 3 y = + + + Page 4 of 8
Type 3: The original curve y f = is unknown, find it. Method Undo the series of transformations one step at a time. At every transformation, replace the variables correctly each time. Method (Advance method, do this only if you understand) y f, y is a point on the curve. Since =, it means ) Apply the series of transformation to the point (, y ). ) Substitute the final point into y g 3) Then make y in terms of. That will be the equation of y f =, the final curve after transformation. =. Eample: SAJC//P/ f undergoes, in succession, the following transformations: The curve A: a translation of 4 units in the positive -direction B: a reflection in the y-ais C: a stretch with scale factor parallel to the -ais The equation of the resulting curve is. Obtain the equation of the original curve + f. Method Let = g. To obtain the original graph we need to work backwards and reverse + each transformation, starting with the last. C : Stretch with scale factor parallel to -ais. Replace with + B : Reflection in the y-ais. Replace with ( ) = + + A : Translation of 4 units in the negative -direction. Replace with +4 ( + 4) 9 + 9 = = + 4 + 7 + 7 Hence the original curve is + 9 + 7. Page 5 of 8
Method Let (, y ) be a point on the original curve y f =. A: a translation of 4 units in the positive -direction + 4, y B: a reflection in the y-ais ( ( + 4 ), y) = ( 4, y) C: a stretch with scale factor parallel to the -ais 4, 8, y The point ( 8, y) lies on the curve ( ) + Hence the original curve is 8 9 + 9 = = 8 + 7 + 7 + 9 + 7. Page 6 of 8
Applying transformation to conics/basic curves Transformation can be applied to conics. You just need to remember the basic equations of the curves and then applied linear transformation for other variations. Basic idea When you replaced the variable with a the curve is translated a units in the positive -ais. Similarly, when you replaced the variable y with y b the the curve is translated b units in the positive b -ais. Note the difference in sign Eample: Circle ( ) ( y ) 4 + + = 5 The basic equation of a circle is + r. For this question, we will first look at + 5. This is a circle with centre at the origin and radius 5. ( ) ( y ) 4 + + = 5 We are replacing with 4. We shifted the circle 4 units in the positive -ais. We are replacing y with y +. We shifted the circle units in the negative y -ais. Therefore ( 4) + ( y + ) = 5 is a circle with centre at ( 4, ) Eample: Hyperbola ( ) ( y ) =. 3 The basic equation of this type of hyperbola is look at y 3 ( ) ( y ) a =. The centre of the hyperbola is and radius 5. y =. For this question, we will first b 3 0,0, the asymptotes are ±. = 3 We are replacing with. We shifted the units in the positive -ais. We are replacing y with y. We shifted the units in the positive y -ais. The centre is now at (, ). The asymptotes are now 3 3 3 3 7 ( y ) = ± ( ) ± ( ) + + or + Page 7 of 8
Eample: ln ( ) The basic equation is ln. The vertical asymptote is 0 = and the -intercept is,0. ( ) ln We are replacing with. We shifted the units in the positive -ais. The new vertical asymptote is = 0 =. 3,0. The new -intercept is Eample: e + 3 The basic equation is y = e. The horizontal asymptote is 0 and the y-intercept is 0,. ( y 3) e + 3 = e. We are replacing y with y 3. We shifted the 3 units in the positive y -ais. The horizontal asymptote is y 3 = 0 3 and the y-intercept is ( 0, 4 ). Page 8 of 8