Module Summary Sheets. C1, Introduction to Advanced Mathematics (Version B reference to new book)

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1 MEI Mathematics i Educatio ad Idustry MEI Structured Mathematics Module Summary Sheets C, Itroductio to Advaced Mathematics (Versio B referece to ew book) Topic : Mathematical Processes ad Laguage Topic : Algebra. Basics. Quadratics Fuctios. Iequalities 4. Idices Topic : Coordiate Geometry. Lies. Curves Topic 4: Polyomials. Maipulatig Polyomials. Biomial Expasios Topic : Curve Sketchig Purchasers have the licece to make multiple copies for use withi a sigle establishmet September, 004 MEI, Oak House, 9 Epsom Cetre, White Horse Busiess Park, Trowbridge, Wiltshire. BA4 0XG. Compay No Eglad ad Wales Registered with the Charity Commissio, umber 089 Tel: Fax:

2 Summary C Topic : Mathematical processes ad laguage Chapter 6 Pages 49- Exercise 6B Q., 7 Chapter 6 Pages 4-7 Exercise 6D Q., Chapter 6 Pages - Exercise 6C Q., Symbols I makig logical deductios we use the symbols A B meas statemet A leads to statemet B A B meas statemet A follows from statemet B A B meas statemet A is equivalet to statemet B A B meas A is sufficiet for B A B meas A is ecessary for B Theorems are geeral statemets. If they are true the they will be true i all circumstaces. To prove a theorem therefore requires a proof that covers all possibilities. Showig the assertio to be true for a specific case is ot good eough. E.g. to prove that the sum of itegers = ( + ) Substitute =, the + = ( + ) = This does ot costitute a proof. E.g. Sum of agles of triagle = 80 o Suppose you draw a triagle ad measure the agles ad they come to 80 o. This fails to costitute a proof o two couts:. You caot measure accurately so you ca oly assert that the sum is approximately 80 o.. It is oly for oe triagle ayway ad ot all triagles. Disprovig a theorem A assertio ca be disproved by a sigle example. E.g. Is prime? For = 0,,, etc, it is prime, but = 4 gives a umber which is ot prime. This sigle couter-example shows that the theorem is ot true. The coverse of a theorem This is a theorem stated the other way roud. If the theorem ca be represeted by A B the the coverse is represeted by A B. E.g. Pythagoras Theorem. If a triagle is right-agled the a + b = c. The coverse is: If a + b = c the the triagle is right-agled. The coverse of a theorem is ofte, but ot always true. E.g. If a quadrilateral has four agles = 90 0 the opposite sides will be equal. The coverse is ot true. i.e. If a quadrilateral has opposite sides equal the it is ot true that the agles are all equal to (e.g. a rhombus.) E.g. Statemet A: The three agles of a triagle are equal Statemet B: The three sides of a triagle are equal Which of the followig are correct? A B, A B, A B. A leads to B because if the agles are equal the the triagle is equilateral ad so all three sides are equal. But also if the sides are equal the the triagle is equilateral ad so the agles are equal. So A B is correct. E.g. Statemet A: x = x Statemet B: x = Which of the followig are correct? A B, A B, A B. A does ot lead to B as this is a icomplete statemet. However, if x =, the x = x. So A B is correct. Note that if B had bee x = 0 or the A B would have bee correct. Mathematical Modellig A situatio i real life ca sometimes be expressed exactly mathematically (e.g. the cost of a kilogram of apples is 47p). Sometimes it caot be doe exactly, or it is ot coveiet to do so. E.g. fittig a fuctio to a curve from observed data. E.g. makig assumptios i order to be able to express it mathematically i a form which ca be used. I either situatio we say that we are makig a mathematical model. Questios i the examiatios (at all levels ad i all strads) will refer to a mathematical model as a mathematical descriptio of a real-life situatio. C; Itroductio to Advaced Mathematics Versio B: page Competece statemets p, p, p

3 Summary C Topic : Algebra ; Basics The Laguage of Algebra Word Example Descriptio Fuctio f(x) = x x + Iput a value, x, output a uique value f(x) or y. A graph could be draw. e.g. whe x =, y or f(x) = 4. Variable x, y, A value which ca vary (take differet values). Costat, c A fixed umber. Expressio x x + A expressio ivolves variables ad costats (there is o = sig) Equatio x x + = 0 A equatio ca be solved for specific values of x. Term x x is a term of the expressio x x + Coefficiet is the coefficiet of the x term i the expressio x x + Idex x is the highest power (idex) of x i the expressio x x + Chapter Pages - Chapter Pages - Exercise A Q. (i),(v), (i), (i), 4(v), (v), 6(v), 7(v) Simplifyig Algebraic Expressios Algebraic expressios ca be simplified as follows: Addig like terms: E.g. x+ y x+ y= x+ y Cacellig commo factors i a fractio: 6xy x E.g. = y y Multiplyig out brackets: E.g. (x + ) = 6x+ By factorisig: E.g. x 6xy = x( x y) E.g. Simplify x+ y x y = x x+ y y = x y E.g. Simplify (x+ y) (4 x y) = 4x+ 6y x+ y = 8x+ 9y E.g. Simplify ( x+ y) ( x+ y) xy x+ y x 6y y = = = xy xy x E.g. Factorise 8x y z 6xyz ( ) H.C.F. of the two terms is 9xyz 9xyz 9xy 4z Chapter Pages 7-9 Exercise B Q. (iii), (x),, Chapter Pages - Exercise C Q., 6, C; Itroductio to Advaced Mathematics Versio B: page Competece statemets a, a, a, a4 Liear Equatios Liear equatios ivolve oly a sigle power of x. A geeral order of approach: Clear fractios. Multiply out brackets. Gather terms & simplify. Divide. Trasposig formulae (Chagig the subject of formulae) Follows the same rules as solvig liear equatios. Clear fractios. Multiply out brackets. Gather terms & simplify. Factorise if ecessary. Divide. x E.g. Solve ( x + ) = : 6 + = x [ ] ( x ) 6x+ 8 + x= 0 7x = [ 7: ] x = 7 E.g. Traspose for x(make xthe subject) i x+ the formula y = x x+ y= y(x ) = x+ x xy y = x+ xy x = + y + y x(y ) = + y x= y

4 Summary C Topic : Algebra ; Quadratic Fuctios Chapter Pages -8 Quadratic fuctios are of the form f(x) = ax + bx + c Their graphs are parabolas. If a > 0 the f(x) has a miimum poit. If a < 0 the f(x) has a maximum poit. E.g. f( x) = x x + E.g. f( x) = x + x+ Chapter Pages 4-8 Exercise D Q. (v), (ii), (i) (ii) Quadratic factorisatio (x a)(x b) = x (a + b)x + ab. Therefore, to factorise a quadratic fuctio you eed to fid the two umbers a ad b such that their sum is the coefficiet of x ad their product is the costat term. E.g. x 8x + has the two umbers ad x 8x + =(x )(x ) E.g. Factorise x + x+ + = ad = x + x+ = x+ x+ E.g. Factorise x x 6 + = ad = 6 x x = x x+ 6 Chapter Pages8-4 Exercise D Q. 6(ii), 7(ii), 8(ii), Quadratic equatios are of the form ax + bx + c = 0 The values of x that satisfy this equatio are called the roots. The solutio of the equatio is the set of all the roots. Graphically this is where the curve y = ax + bx + c cuts the x-axis. A quadratic equatio ca have 0, or roots. Quadratic equatios may be solved by factorisig ad puttig each factor equal to zero, or by usig the formula: b± b 4ac x = a If the discrimiat, b 4ac is egative the there is o square root ad so there are o roots. If b 4ac = 0 the the two roots are coicidet. E.g. Solve the equatio x + x = 0 x + x = 0 ( x )( x ) + = 0 = 0 x = + = 0 x = ( x ) ( x ) 4 E.g. Solve the equatio x = x x x 4 = 0 (equatio does ot factorise) ± x = 4 x = ( ± ) x =. or Chapter Page 4 Exercise D Q. 8 (iii), (iv) Completig the square is the process of puttig a quadratic expressio i the form a(x p) + q. Hece we ca say that the fuctio will have a miimum (or maximum) poit at ( p, q). A alterative method is to that show is to compare coefficiets. ( x p) + q x px+ p + q Therefore : x 6x ( x ) 9 ( x ) 4 E.g. Complete the square for x 6x. Hece state the coordiates of the miimum poit ad the equatio of the axis of symmetry. x 6x x 9 x 4 Has mi. poit at (, 4), Axis of symmetry at x = C; Itroductio to Advaced Mathematics Versio B: page 4 Competece statemets a, a6, a7 x = (, 4)

5 Summary C Topic : Algebra ; Simultaeous Equatios ad Iequalities Chapter Pages 8-0 Exercise E Q. (ii), Chapter Page 0 Exercise E Q. (ii), 6 Chapter 4 Pages -4 Exercise 4A Q. (i), (ii) Chapter 4 Pages 4- Exercise 4A Q. (i), (v) Liear simultaeous equatios Equatios i two (or more) variables that are true simultaeously are kow as simultaeous equatios. Two liear equatios i two variables may be solved by: Substitutio (oe variable is made the subject of oe Equatio ad substituted i the other) Elimiatio (both equatios are maipulated so that a coefficiet of oe variable is the same. That variable is the elimiated.) Graphical solutio see page 7 Simultaeous equatios whe oe is oliear Make oe variable the subject of the liear equatio ad substitute i the other equatio. I this uit the result will be a quadratic equatio. Iequalities ca be solved usig the same rules as for equatios except: Whe multiplyig by a egative umber the directio of the iequality sig reverses. Care eeds to be take to maitai the iequality sig. You should be familiar with iequality diagrams. Quadratic iequalities You are advised always to sketch a graph. For the iequality, ax + bx + c > 0, sketch the curve y= ax + bx+ c. Fid where the curve crosses the x-axis (say at a ad b) ad this gives oe of the two results, a< x< b or x< a ad x> b. There are two ways of fidig the solutio. (i) Sketch the graph accurately eough to determie where it crosses the x-axis. (ii) Factorise the quadratic expressio. This gives a product of two factors. For a greater tha iequality both factors must be positive. For a less tha iequality oe must be positive ad the other egative. C; Itroductio to Advaced Mathematics Versio B: page Competece statemets a, a, a4, a8, a9 E.g. Fid the poit of itersectio of the lies x y = 7 ad x + 4y =. x y = 7 (i) x + 4 y = (ii) (i)x4 8x y = 8 (iii) (ii)x 9x + y = 6 (iv) (iii) + (iv) 7x = 4 x = Sub i (ii) : 6 + 4y = y = E.g. Fid the poit of itersectio of the lies x y = ad 4x + y = 7 x y= y= x Sub. ito 4x+ y= 7 4x+ ( x ) = 7 9x 0= 7 9x= 7 x= y= E.g. Solve simultaeously the followig equatios y = x ad y = x x+ Substitute for y: x = x x+ x x+ 4 = 0 ( x )( x ) 4 = 0 x =, 4 (,), (4, 7) E.g. Solve the iequality 6 x < 6 x < 6 < x or x < 6 < x or x < < x or x > E. g. Solve 0 [ ] x x + : x + x 0 (This factorises) (x )( x + ) 0 x, x (See graph) x = x = /

6 Summary C Topic : Algebra 4; Idices Chapter Pages 7-9 Exercise A Q. (iii), (iii), (v), (iii), (iv) Chapter Page 8 Exercise A Q. 4(i), (iii) Surds ad irratioal umbers The square root of a positive iteger is either a iteger or a irratioal umber. A irratioal umber is oe that, whe writte as a decimal, either termiates or recurs, ad so caot be writte as a fractio. A umber which icludes such a value is called a surd. Note that this is a exact value, while ay decimal approximatio (e.g. +.7 =.7) is oly a approximatio. Ay questio that requires a exact aswer is almost certaily goig to ivolve surds. Take care whe you see the requiremet for a exact aswer i a questio. Ratioalisig the deomiator Whe a surd appears i the deomiator, the factio may be "ratioalised" by multiplyig E.g. 4 = E.g. E.g. + top ad bottom by the surd so that the deomiator becomes a ratioal umber. E.g. = = ( + )( ) = 4 = (Differece of squares) has a irratioal deomiator which ca be ( + ) "ratioalised" by multiplyig by ( ). ( ) e.g. = = ( ) ( + ) ( + ) ( ) E.g. + = = + = E.g. Fid the roots of the equatio x + 4x = 0 exactly. 4 ± ± 76 4 ± 9 x = = = = ( 9 ) ad ( + 9 ) N.B. Use of calculators here to fid a value of the square root will result i a approximate aswer. Note the requiremet for a exact aswer. Remember that calculators are ot allowed i the examiatio so you will ot be asked for a approximate aswer. E.g. Simplify + ( ) = = = E.g. Simplify + + = + ( + ) = = Chapter Pages 0- Exercise B Q.4(iv), 6(i), 7(iv), 8(ii), 9(iii) Idices The Laws of idices m m+. a a = a m m. a a = a m. ( a ) = a 4. m= i a = C; Itroductio to Advaced Mathematics Versio B: page 6 Competece statemets a, a4, a, a6 m 0. m= 0 i ad usig 4 a = a 6. = m= i a= a 7. more geerally: m a= a m I the umber a m, a is called the base ad m is called the power or idex. E.g. = = = - 6 =, =, = ; 64 = 4 9 = 9 = = 7;also 9 = 9 = 79 = 7 N.B. Mixed Bases, for istace, caot be evaluated usig the laws of idices because the bases are ot the same. = 8 9 = 7 Note, however, that = 6 = 6

7 Summary C Topic : Coordiate Geometry ; lies Chapter Pages 4-40 Exercise A Q. (i), (iv), 7 Chapter Pages 4- Exercise B Q. (v), (x) Exercise C Q.(v), (v), 4(v), 6, 0 Exercise B Q. (ix), (xv) Properties of lies For the lie defied by the poits, ad, y y Gradiet = x x If a lie has gradiet m Parallel lies also have gradiet m Perpedicular lies have gradiet - m Legth = y y + x x x + x y + y Mid poit =, Equatios of lies The equatio of a lie passig through ( x, y) with gradiet mis y y = m x x ( ) ( x y ) ( x y ) Gradiet - itercept form of a straight lie, with gradiet m, itercept o yaxis c, is y=mx+c The form px + qy + r = 0 is usually used to esure that fractios are ot icluded. p Rearragig will give gradiet = ad itercept o the q r r axes,0 ad 0,. p q Sketchig straight lie graphs The sketch should show whether the slope is positive or egative ad where the lie cuts the axis. Do't worry about puttig scales o the axes Lie cuts the y axis where x = 0. Lie cuts the x axis where y = 0. E.g. Fid the gradiet, legth & mid poit of the lie betwee the poits (,) & (4, ). 4 gradiet = = = 4 ( ) 6 ( ) ( ) legth = + 4 = 6 +6 = +4 +( ) mid poit =, =, E.g. Fid the equatio of a lie with gradiet 4, passig through the poit (, ). y+ = 4 x 4x+ y = E.g. Fid the equatio of a lie passig through (, ) ad perpedicular to the lie x + y =. x+ y =: y = x + (puttig ito gradiet - itercept form) has gradiet So perpedicular gradiet is & required eq is : y = ( x+) x y = 9 E.g. Sketch the lie x y = yaxis ( x = 0 ): y = y = xaxis ( y = 0 ): x = x = y x Alteratively: y = x cuts y axis at Gradiet (+ve) = E.g. Fid the poit of itersectio of the lies 4x + y =, x y = 9 Plot the lies. Chapter Pages -6 Exercise D Q., Itersectio of lies Two o - parallel lies meet at a poit. The coordiates of this poit satisfy both equatios simultaeously. Solve as simultaeous equatios. C; Itroductio to Advaced Mathematics Versio B: page 7 Competece statemets g, g, g, g4, g, g6, g7, g8 They itersect at (, ).

8 Summary C Topic : Coordiate Geometry ; curves Chapter Pages 6-66 Exercise E Q. (ii), (ii),8 Chapter Page 6 Chapter Pages Exercise F Q.4, 6 Chapter Pages 7-7 For the geeral shape of curves, see Topic The Circle The circle with equatio (x a) + (y b) = r has cetre (a, b) ad radius r. The circle with equatio x + y hx ky + c = 0 simplifies to (x h) + (y k) = h + k c, givig the radius, r, as r = h + k c A lie, i geeral, cuts a circle i two poits. Substitutio for y from the equatio of the lie ito the equatio of the circle will give a quadratic equatio i x. The equatio may have 0, or roots. A chord (or diameter) will give two poits; a taget will have coicidet roots. If the lie does ot cut the circle the this quadratic equatio will have o roots. The Circle Theorems The agle at the cetre is twice the agle at the circumferece Agles i the same segmet are equal The agle subteded by a diameter is a right agle The agle betwee a taget ad a chord at a poit is equal to the agle i the alterate segmet. (ot icluded i the specificatio.) Itersectio of a lie ad a curve A lie may cut a curve i distict poits, or it may touch it. A lie that touches a curve is called a Taget. It is also possible that the lie does ot cut the curve. The process of fidig the poits is usually as follows: Write the equatio of the lie i the form y =. Substitute this expressio for x ito the equatio of the curve. Solve the resultig equatio i x. Substitute the values of x that satisfy this equatio ito the equatio for the lie to give the y values of the poits of itersectio. If the lie does ot meet the curve the there will be o roots to the equatio. If the lie touches the curve the there will be two coicidet roots to the equatio. Itersectio of curves The itersectio(s) of two curves ca be foud at this level oly if the above procedure ca be adopted - i.e. if oe of the variables ca be made the subject of oe equatio ad the substituted ito the other to give a equatio (i either x or y) which ca the be solved. Two curves ca itersect i ay umber of poits; the resultig polyomial equatio i x will be of the order of the umber of itersectios I.e. if two curves cut i two poits (e.g. two circles) the the equatio to be solved will be quadratic. E.g. Fid the equatio of the circle with cetre (,) ad radius. (x ) + (y ) = gives x + y x 4y = 0 E.g. Fid the radius ad cetre of the circle x + y 6x + 4y + 9 = 0 The equatio simplifies to (x ) + (y + ) = = 4. Thus the cetre is (, ) ad the radius is. E.g. Fid where the lie y = x + cuts the curve y = x x. Sub. for y: x + = x x Gives x 4x = 0 (x )(x + ) = 0 x = or Sub. i lie: y = 0 or 4 The lie cuts the curve at (, 4) ad (, 0). E.g. Fid where the lie y = 4x 6 cuts the curve y = x + x. Sub. for y: 4x 6 = x + x Gives x x + = 0 (x ) = 0 x = twice. Sub. i lie: y =. E.g. Fid where the curves y = x + ad x + y = itersect. st equatio gives x = y Substitute ito d equatio: y + y = y + y = 0 (y )(y + ) = 0 y =, y = has o solutio for x. y = gives x = 4 x = ± itersectios at (,) ad (,) C; Itroductio to Advaced Mathematics Versio B: page 8 Competece statemets g9, g0, g, g, g, g4

9 Summary C Topic 4: Polyomials ; Maipulatig Polyomials Chapter Pages Chapter Pages 78-8 Exercise A Q., 7, 6 Polyomials are expressios of the form ax +bx +...px+ q is a positive iteger; a,b,...p,q are real umbers. Addig ad subtractig polyomials Remove brackets, collect like terms Multiplyig polyomials Careful settig out ca facilitate the gatherig together of like terms. Dividig polyomials Set out as for a arithmetic divisio. E.g. x + x 4 x is a polyomial of degree. ( x x + x ) ( x + x x+ ) E.g. 4 4 = x x + 4x x + x + x 4 = x + x + x 6 E.g. Divide x + x 6x+ 4 by x E.g. ( x x+ )( x + x x + x 4 ) 4 = x + x x ( x ) x + x 6x+ 4 6x x + x x x x 6x + 4x + x 0 x x 4 4x + 4 = x x 4x + 7x 0 4x + 4 Chapter Pages 8-86 Exercise B Q., 4, 6 Sketchig Polyomial Fuctios Kow the basic shapes. Fid where the curve cuts the y-axis ( x= 0) Fid where the curve cuts the x-axis ( y= 0) (solve the resultig equatio by factorisig where possible) Coordiates of turi g poits ca be foud by differetiatio. Iequalities Ca be solved by sketchig a graph. E.g. Sketch the curve y = x(x ). The curve is a cubic, so has two turig poits. As x gets large so does y. Whe x = 0, y = 0 Whe y = 0, x = 0 ad (twice) Chapter Pages 89-9 Chapter Pages 9-9 Exercise C Q., The Factor Theorem x a is a factor of f(x) f(a) = 0 If a polyomial ca be factorised the the factors ca be foud by trial of various umbers, a. E.g. Suppose f(x) = (x a) p(x) where p(x) is a polyomial of order oe less tha the order of f(x). The if x = a the value of f(x) is 0 whatever the value of p(x). The Remaider Theorem If f(x) is divided by (x a) the the remaider is f(a). Note that whe the remaider is 0 (x a) is a factor, so this is a extesio of the Factor Theorem. E.g. Fid the remaider whe x + x 4x 4 is divided by ( x ), ( x + ) f 4 4 f 6 The remaider is 6 f 0 The remaider is 0 (so is a factor. ( x) = x + x x () = ( ) = ( x + ) E.g. Solve, by factorisig, x + x 4x 4 = 0 f ( x) = x + x 4x 4 f () = 6 (So x is ot a factor) f ( ) = 0 x + is a factor f = 0 x is a factor The third factor is x +, by cosiderig the last term. So f ( x) = ( x+ )( x )( x+ ) x =,, E.g. ax x + 4 is divisible by x+. Fid a. f - = a +4= a= 0 C; Itroductio to Advaced Mathematics Versio B: page 9 Competece statemets f, f, f, f4, f, f6

10 Summary C Topic 4: Polyomials ; Biomial Expasio Chapter Pages 08-0 Exercise F Q. (iii),(v) Biomial expasios Whe a liear expressio of the form (a + x) is multiplied by itself a umber of times the a polyomial results. This polyomial ca be writte dow otig the followig: For (a + x) Each term is of the same power, startig with a with decreasig powers of a ad icreasig powers of x. Each term has associated with it a coefficiet. For (a + x) each term is positive for positive a. For (a x) the terms are alteratig i sig for positive a. For small positive the coefficiets ca be foud from Pascal s Triagle (See Studets Hadbook for a larger versio) E.g. Simplify ( + x) + ( x). Hece fid ( + x) + ( x) = + x + x + x + x + x x = + 6x Put x = 0.0 Gives = + 6(0.000) =.0006 E.g. Expad ( + x) 4. ( + x) 4 = x + 6..x +4. x +x 4 = 6 + x + 4x +8x + x 4. E.g. ext two lies are: Chapter Pages 0- Exercise F Q. (iii),(iv) Chapter Page 4 The formula for a biomial coefficiet The formula for the coefficiet of the rth term of ( a + x) is give by! = Cr = where! = ( )( )... r ( r)! r! ( )( )...( r+ ) r = r E.g. = =.. The expasios are therefore as follows: x x x x ( ) ( )( ) = + x + x + x a+ b = a + a b+ a b +... b + = Relatioships betwee biomial coefficiets r r + r r+ r+ C = C C + C = C! ( + )! Cr + Cr+ = + r!( r)! ( r+ )!( r )!! +! = + = r!( r )! r r+ ( r+ )!( r)! + + ( + )! + = = Cr+ ( r+ )!( r)! Substitute x= ito ( + x) to give C + C C = 0 ( r r) E.g = = = = = = = =...4. E.g. Fid the coefficiet of x i ( x). ( x) =..x + 0..x 0. x +. Coefficiet is 80. E.g. Fid the term idepedet of x i the expasio of x. x x = ( x) 4.( x) + 6( x) +... x x x Term idepedet of x is 6( x) x = = E.g. C =, C =, C =, C = 0 C + C + C + C = = 8= 0 Note also that the sum of the lies of Pascal s Triagle are the powers of. C; Itroductio to Advaced Mathematics Versio B: page 0 Competece statemets f7, f8, f9

11 Summary C Topic : Curve sketchig Plottig ad sketchig A curve ca be plotted poit by poit from its equatio. The geeral shape may be missed, however, if the rage of values of x is ot large eough or ot small eough. A curve ca be sketched, ad so the geeral shape discovered The geeral shape of stadard curves should be kow. y = kx y = kx Chapter Pages 60-6 y = kx Chapter Pages 8-87 k y = x Chapter Pages -9 Chapter Pages Fidig the miimum value of a quadratic by completig the square. (See page 4 for this topic). The quadratic fuctio y = ax + bx + c is a parabola of the followig forms E.g. Fid the miimum value of x 4x + 9. (x ) = x 4x + 4 x 4x + 9 = x 4x = (x ) + So miimum value is whe x =. Exercise D Q. (v), (vi) a > 0 a < 0 Chapter Pages 0-0 Exercise E Q. (i), (vi) Usig trasformatios to sketch the curves of fuctios y= f ( x a) is the curve y= f ( x) traslated a uits i the +ve x directio. y= f ( x) + a is the curve y= f ( x) traslated a uits i the +ve y directio. E.g. f:x x g:x (x + ) i.e. g(x) = f(x + ) g f y = x y = (x ) y = x y = x C; Itroductio to Advaced Mathematics Versio B: page Competece statemets C, C, C, C4