20 For free distribution

By studying this lesson you will be able to• findthegradientofthegraphofastraightline,• drawthegraphofafunctionoftheform y = ax2 + b.

Graph of a function of the form y = mx + c

Thegraphofafunctionoftheform y = mx + c isastraightline.Thecoefficientofx,whichism,representsthegradientofthelineandtheconstanttermcrepresentstheinterceptofthegraph.

Review Exercise

1.Writedownthegradientandinterceptofthestraightlinerepresentedbyeachofthefollowingequations.

21.1 Geometrical description of the gradient of a straight lineWedefined thecoefficient m of x in theequation y = mx + c as thegradientofthestraightline.Nowbyconsideringanexample,letusseehowthevalueofm isrepresentedgeometrically.Todothis,letusconsiderthestraightlinegivenbyy = 2x + 1. Letususethefollowingtableofvaluestodrawitsgraph.

x – 2 0 2y (= 2x + 1) – 3 1 5

Graphs21

21For free distribution

Letusmarkanythreepointsonthestraightline.Forexample,letustakethethreepointsasA(0,1),B(2,5)andC(5,11).

^5" 11&

^2" 5&

−6

1

1

−5

−5

2

2

−4

−4

3

3

−3

−3

4

4

−2

−2

5

5−1

−1

6789

10

11

0

^2" 1&

^5" 5&

A

B

Cy

x

Q

P^0" 1&

FirstletusconsiderthepointsAandB.LetusdrawalinefromA,paralleltothex –axis,andalinefromB,paralleltothey –axis,andnamethepointofintersectionoftheselinesasP.ItisclearthatthecoordinatesofthepointPare(2,1).Also,lengthof AP =2–0 =2 lengthof BP =5–1 =4NowforthepointsAandB, Verticaldistance

Horizontaldistance = = =BP 4AP 2 2 .

Wealreadyknowthegradiantofthestraightliney=2x+1is2.

Thequotient VerticalDistanceHorizontaldistance

forthepointAandBisalso2.

Nowletusconsideranothercase.AsthesecondcaseletusconsiderthepointsBandC.LetusdrawalinefromB,paralleltothex-axis,andalinefromC,paralleltothey-axisandnamethepointofintersectionofthesetwolinesasQ. ThenthecoordinatesofQare(5,5). Lengthof BQ=5–2 =3

22 For free distribution

Lengthof CQ =11–5 =6

Now,forthepointsB andC, VerticaldistanceHorizontaldistance = = =CQ 6

BQ 3 2 .

Inbothinstances,theratiooftheverticaldistancetothehorizontaldistancebetweenthetwopointsunderconsiderationisthegradient2ofthestraightline.

Accordingly,letusdevelopaformulatofindthegradientofastraightlineusingitsgraph.Letusconsideranystraightlinewithequationy = mx + c.

x

y

−6

1

1

−5

−5

2

2

−4

−4

3

3

−3

−3

4

4

−2

−2

5

5−1

−1

67

0

A

B

(x1,y1)

(x2, y2)

LetusconsideranytwopointsA(x1,y1)andB (x2,y2)onthestraightline.Sincethesetwopointslieonthestraightline,

The gradient of the straight line

From and

23For free distribution

Example 1

Thecoordinatesof twopointsona straight lineare (3,10)and (2,6).Find thegradientofthestraightline.

Gradient of the straight line

Example 2

The coordinates of two points on a straight line are (6, 3) and (2, 5). Find thegradientofthestraightline.

Gradient of the straight line

Example 3

Find thegradientof the straight line thatpasses through thepoints (–2,4)and(1,–2).

Gradient of the straight line

24 For free distribution

Exercise 21.11.Calculatethegradientofthestraightlinewhichpassesthrougheachpairofpoints.(i) (4,6) (2,2) (ii) (6,2)(4,3) (iii) (1,–2)(0,7) (iv)(–2,–3)(2,5)(v) (4,5)(–8, –4) (vi) (6,–4)(2,2) (vii) (1,–4) (–2,–7) (viii) (4,6)(–2,–9)

21.2 Finding the equation of a straight line when the intercept of its graph and the coordinates of a point on the graph are given

Example 1

Theinterceptofthegraphofastraightlineis3.Thecoordinatesofapointonthegraphis(2,7).Writetheequationofthestraightline.

Theequationofastraightlinegraphwithgradientmandinterceptcis y = mx + c.

Bysubstituing thevalueof the interceptand thecoordinatesof thepointon thegraphintotheequationofthefunctionweobtain, y = mx + c

By substituting c = 3 and m = 2 into the equation we obtain y = 2x + 3.

x

Exercise 21.21. Foreachgraphwith thegiven interceptandpassing through thegivenpoint,writedowntheequationofthecorrespondingfunction.

(i)Intercept = 1and(3,10) (ii) Intercept = 2and(3,3)(iii) Intercept = 5and (2, 1) (iv) Intercept = 0and(3,12)(v) Intercept = – 4and(3,8) (vi) Intercept = –5and(–2,–9)

25For free distribution

21.3 Finding the equation of a straight line which passes through two given points

Letusfindtheequationofthestraightlinewhichpassesthroughthepoints(1,7)and(3,15).Toobtaintheequation,letusfindthegradientandtheinterceptofthegraph.Letusfirstfindthegradientofthestraightline,usingthecoordinates(1,7)and(3,15)ofthetwopointsontheline.

Letussubstitutethevalueofmandthecoordinatesofoneofthepointsintotheequation y = mx + c. Therebywecanfindthevalueofc.

×1

m = 4 and c = 3Thegradientofthegraphis4andtheinterceptis3.Therefore,therequiredequationisy=4x+3.

Example 1

Findtheequationofthestraightlinepassingthroughthepoints(4,3)and(2,–1).

26 For free distribution

Nowletussubstitutethecoordinatesofthepoint(2,–1)andthegradientintotheequation y = mx + c.

The equation of the straight line is

Exercise 21.31.Findtheequationofeachofthestraightlinegraphsthatpassesthroughthegivenpoints.

(i) ^1" 7& ^2" 10& (ii) ^3" –1& ^–2" 9& (iii) ^4" 3& ^8" 4& (iv) ^2" –5& ^–2" 7&

(v) ^–1" –8& ^3" 12& (vi) ^–5" 1& ^10" –5& (vii) (viii) ^2" 2& ^0" –4&

21.4 Graphs of functions of the form y = ax2

Now let us identify several basic properties of graphs of functions of the formy=ax2.Hereaisanon-zeronumber.Hereyisdefinedasthefunction.ycanbeconsideredasafunctionwhichisdeterminedbyax2.

Firstletusdrawthegraphofy = x2.Todothis,letusproceedaccordingtothefollowingsteps.

Step 1Preparingatableofvaluestofindtheyvaluescorrespondingtogivenxvaluesofthefunction. y = x2

x –3 –2 –1 0 1 2 3x2 9 4 1 0 1 4 9y 9 4 1 0 1 4 9

Using the tableofvalues, letusobtain thecoordinatesof thepoints required todrawthegraphofthefunction.(–3,9),(–2,4),(–1,1),(0,0),(1,1),(2,4),(3,9)

Step 2PreparingaCartesiancoordinateplanetomarkthecoordinatesthatwereobtained.

27For free distribution

Themaximumvaluethatthexcoordinatetakesinthepairsthatwereobtainedis+3andtheminimumvalueis–3.Themaximumvaluethattheycoordinatetakesis9andtheminimumvalueis0.Letusdrawthexandyaxesaccordingtoasuitablescale,onthepieceofpaperthatistobeusedtodrawthegraph,sothatthex-axiscanbecalibratedfrom–3to3andthey-axiscanbecalibratedfrom0to9.

Step 3Drawingthegraphofthefunction.Letusmarkthesevenpointsonthecoordinateplanethathasbeenprepared.Nextletusjointhepointsthathavebeenmarkedsothatasmoothcurveisobtained.Thissmoothcurveisthegraphofthefunctiony = x2.

y

x0 1

1

–1–1–2–3–4

2

3

4

5

6

7

2

–2

3 4

8

9

y=x2

Thecurvethatisobtainedasthegraphofafunctionoftheformy = ax2isdefinedasaparabola.Letusidentifyseveralpropertiesofthegraphofthefunctiony = x2byconsideringthegraphthatwasdrawn.Forthefunctiony = x2,• thegraphissymmetricaboutthey-axis.Therefore, they-axisis theaxisofsymmetryofthegraphandtheequationoftheaxisofsymmetryisx=0.

•when thevalueofx increasesnegatively (i.e., –3 to0) the functiondecreasespositivelyandwhenthevalueofxincreasespositively(i.e.,0to+3)thefunctionincreasespositively.

28 For free distribution

Toidentifythecommonpropertiesofthegraphsoffunctionsoftheform y = ax2

where a > 0,letusdrawthegraphsofthefunctions y = x2, y = 3x2 and y = onthesamecoordinateplane. y = 3x2 y =

x −2 −1 0 1 2x2 4 1 0 1 4

x2 2

0 2

y 2

0 2

x −2 −1 0 1 2

x2 4 1 0 1 4

3x2 12 3 0 3 12

y 12 3 0 3 12

(–2,12),(–1,3),(0,0),(1,3),(2,12)(–2,2),(–1, ),(0,0),(1, ),(2,2)

y

x0 1

2

–1–2–3–4

4

6

8

10

12

2

–23 4

y = 3

x2

y =

y=x2

Letusidentifyseveralcommonpropertiesofthegraphsoffunctionsoftheformy = ax2,wherea>0,byconsideringtheabovegraphs. •Thegraphisaparabolawithaminimumpoint. •Thecoordinatesoftheminimumpointare(0,0). •Thegraphissymmetricaboutthey-axis. •Theequationoftheaxisofsymmetryisx=0.

29For free distribution

•Theminimumvalueofthefunction(i,e.,valueofy)is0. •Thefunctiondecreaseswhenthevalueofxincreasesnegatively(increasingalongnegativevalues)andreachestheminimumvalueatx=0. •Thefunctionincreasesfrom0whenthevalueofxincreasespositively(increasingalongpositivevalues).

Toidentifythecommonpropertiesofgraphsoffunctionsoftheform y = ax2 wherea < 0,letusdrawthegraphsofthefunctions , and onthesamecoordinateplane.

y = −x2

x –3 –2 –1 0 1 2 3 x2 9 4 1 0 1 4 9 –x2 –9 –4 –1 0 –1 –4 –9

y –9 –4 –1 0 –1 –4 –9

^–3" –9& ^–2" –4& ^–1" –1& ^0" 0& ^1" –1& ^2"–4& ^3"–9&

y = –2x2

x –2 –1 0 1 2x2 4 1 0 1 4

–x2 –8 –2 0 –2 –8 y –8 –2 0 –2 –8

^–2" –8& ^–1" –2& ^0" 0& ^1" –2& ^2" –8&

x –4 –2 0 2 4 x2 16 4 0 4 16– x2 –8 –2 0 –2 –8

y –8 –2 0 –2 –8

^–4" –8&" ^–2" –2&" ^0" 0&" ^2" –2&" ^4" –8&

30 For free distribution

y

x1–1–2–3–4–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

–12

1

2 3 4

y =

−x2

y =

–2x

2

0

Let us identify the common properties of the graphs of functions of the formy=ax2,whenaisnegative(a<0),byconsideringtheabovegraphs. •Thegraphisaparabolawithamaximumpoint. •Thecoordinatesofthemaximumpointare(0,0). •Themaximumvalueofthefunctionis0. •Thegraphissymmetricaboutthey-axis. •Theequationoftheaxisofsymmetryisx =0. •Thefunctionisincreasingwhen x isnegativelyincreasingandreachesthe

maximumpointwhenx=0. •Thefunctionisdecreasingwhen x ispositivelyincreasing.

Letusidentifythebasicpropertiesofthegraphsoffunctionsoftheformy = ax2,byconsideringthegraphsthathavebeendrawn.Hereaisanynon-zerovalue.

31For free distribution

Forfunctionsoftheform y = ax2, •thegraphisaparabola.

•thegraphissymmetricaboutthe y - axis. Therefore,theequationoftheaxisofsymmetryofthegraphis x = 0.• thecoordinatesoftheturningpoint(i.e.,themaximumorminimumpoint)ofthegraphare(0,0).

•whenthecoefficientofxtakesa“positive”value,thegraphisaparabolawithaminimumpoint.•whenthecoefficientofxtakesa“negative”value,thegraphisaparabolawithamaximumpoint.

Example 1

Byexaminingthefunction,writedownforthegraphofthefunction "

(i)theequationoftheaxisofsymmetry,(ii)thecoordinatesoftheturningpoint,(iii)whethertheturningpointisamaximumoraminimumpoint.

(i)Theequationoftheaxisofsymmetryisx=0.(ii)Thecoordinatesoftheturningpointare(0,0).(iii)Sincethecoefficientofx2inthefunctionisapositivevalue,thegraphhasa

minimumpoint.

Example 2

Byexaminingthefunction,writedownforthegraphofthefunctiony=–4x2,(i)theequationoftheaxisofsymmetry,(ii)thecoordinatesoftheturningpoint,(iii)whethertheturningpointisamaximumoraminimumpoint.

Sincethefunctionisoftheformy = ax2,(i)theequationoftheaxisofsymmetryisx=0.(ii)thecoordinatesoftheturningpointare(0,0).(iii)sincethecoefficientofx2inthefunctionisanegativevalue,thegraphhasa

maximumpoint.

32 For free distribution

Exercise 21.41.Completethefollowingtablebyexaminingthefunction

Function Coordinates of the turning point

Minimum value of y

Maximum value of y

Equation of the axis of symmetry

2.Incompletetablesofvaluespreparedtodrawthegraphsofthefunctionsy=13 x2and aregivenbelow.

x – 6 – 3 0 3 6

y 12 0 3x – 4 – 2 0 2 4

y – 4 – 1 0

y x 213

=

(i)Completethetablesanddrawthegraphsseparately.(ii)Forthefunctions,writedown

(a)theequationoftheaxisofsymmetryofthegraph,(b)thecoordinatesoftheturningpointofthegraph,(c)themaximumorminimumvalue.

3.(i)Usingvalues ofx such that 3 3x− , prepare a suitable table of values todrawthegraphsofthefunctionsy=2x2,y=4x2,y=–13 x

2andy=–3x2. (ii)Drawthegraphsonasuitablecoordinateplane.(iii)Foreachofthegraphswritedown (a)theequationoftheaxisofsymmetry (b)thecoordinatesoftheturningpoint (c)themaximumorminimumvalueofthefunction.

33For free distribution

21.5 Graph of a function of the form y = ax2 + b Letusdrawthegraphofthefunction y = x2+3 toidentifyseveralbasicpropertiesofthegraphofafunctionoftheform y = ax2 + b (Here a ≠ 0).

x –3 –2 –1 0 1 2 3 x2 9 4 1 0 1 4 9 +3 +3 +3 +3 +3 +3 +3 +3 y 12 7 4 3 4 7 12

y=x2 + 3

–3–4 –2 –1 1 2

2

0

6

10

4

8

12

3 4–2

y

x

y=x2+3

Thegraphofthefunction y = x2 + 3 isaparabolawithaminimumpoint.Forthefunctiony = x2 + 3, •theequationoftheaxisofsymmetryofthegraphisx=0. •thegraphhasaminimumpoint,withcoordinates(0,3). •theminimumvalueoftheycoordinatesofthepointsonthegraphis3.Therefore,theminimumvalueofthefunctionis3.Letusdrawthegraphofthefunction y = x2 –2 toidentifythepropertiesofthegraphofafunctionoftheform y = ax2 + b whenthevalueofbisnegative. y = x2 –2

x –3 –2 –1 0 1 2 3 x2 9 4 1 0 1 4 9 –2 –2 –2 –2 –2 –2 –2 –2 y 7 2 –1 –2 –1 2 7

^–3" 7& ^–2" 2& ^1" –1& ^0" –2& ^1" –1& ^2" 2& ^3" 7&

34 For free distribution

–3–4 –2 –1 1 2

2

6

0

4

8

3 4–2

–4

y

x

Thegraphofthefunction y = x2–2 isaparabolawithaminimumvalue.Forthefunctiony = x2–2, •theequationoftheaxisofsymmetryis x = 0. •thecoordinatesoftheturningpointare (0, –2).•theminimumvalueoftheycoordinatesofthepointsonthegraphofthe

functionis –2. Therefore,theminimumvalueofthefunctionis –2.Letusdrawthegraphofthefunctiony=–2x2+3to identifythepropertiesofthegraphofafunctionoftheformy = ax2+b.Whenthevalueofaisnegative. y = –2x2+3

x –3 –2 –1 0 1 2 3 x2 9 4 1 0 1 4 9–2x2 –18 –8 –2 0 –2 –8 –18 + 3 +3 +3 +3 +3 +3 +3 +3 y –15 –5 +1 +3 +1 –5 –15

^–3" –15& ^–2" –5& ^–1" 1& ^0" 3& ^1" 1& ^2" –5& ^3" –15&

–2 –1 1 2

2

0

4

3 4–2

–8

y

x–3–4

–4

–6

–10

–12

–14

–16

y =

–2x2+3

35For free distribution

Thegraphofthefunctiony=–2x2+3isaparabolawithamaximumpoint.Forthefunctiony=–2x2+3, •theequationoftheaxisofsymmetryofthegraphisx=0. •thecoordinatesoftheturningpointofthegraphare(0,3). • the maximum value of the points on the graph of the function is 3.Therefore,themaximumvalueofthefunctionis3.

Letusidentifyseveralcommonpropertiesofthegraphsoffunctionsoftheformy =ax2+b,byexaminingthegraphsthatweredrawnoffunctionsofthisform.

Thegraphofafunctionoftheformy=ax2+b,•isaparabolawithaminimumpointwhenaisapositivevalue.•isaparabolawithamaximumpointwhenaisanegativevalue.•theequationoftheaxisofsymmetryofthegraphisx=0.•thecoordinatesofthemaximumorminimumpoint(turningpoint)is(0,b). •themaximumorminimumvalueofthefunctionisb.

Example 1

Writedownforthegraphofthefunction y = 3x2– 5,(i)theequationoftheaxisofsymmetry,(ii)thecoordinatesoftheturningpoint,(iii)themaximumorminimumvalueof thefunction.(i)Sincethegraphsoffunctionsoftheform y = ax2+ b areparabolaswhich

aresymmetricaboutthe y - axis, theequationoftheaxisofsymmetryofthegraphofthefunction y = 3x2–5 is x = 0.

(ii) Sincethecoordinatesoftheturningpointofthegraphsoffunctionsoftheform y = ax2 + b are(0,b),thecoordinatesoftheturningpointofthegraphofthefunction y = 3x2– 5 are(0, –5).

(iii)Since thecoefficientofx2 in y = 3x2–5 ispositive, thegraphhasaminimumvalue.Theminimumvalueofthefunctionis –5.

Example 2

Writedownforthefunction y = 4 –2x2, (i)theequationoftheaxisofsymmetryofthegraph, (ii)thecoordinatesoftheturningpointofthegraph,(iii)themaximumorminimumvalueof the function. (i)Theequationoftheaxisofsymmetryofthegraphof y = 4–2x2, is x = 0. (ii)Thecoordinatesoftheturningpointare(0,4). (iii)Since thecoefficientofx2 iny=4–2x2 isnegative, thegraphhasa

maximumvalue.Themaximumvalueofthefunctionis4.

36 For free distribution

Exercise 21.51. Withoutdrawingthegraphsofthefunctionsoftheformy = ax2+bgivenbelow,completethefollowingtable.

Function Equation of the axis of symmetry

Coordinates of the turning point of the graph

Whether the graph has a maximum or minimum value

The maximum or minimum value of the function

2. Incomplete tables of values prepared to draw the graphs of the functionsy=2x2–4andy=–x2+5aregivenbelow.

x –3 –2 –1 0 1 2 3 y –4 ___ +4 +5 ___ +1 –4

x –2 –1 0 1 2 y 4 ___ ___ –2 4

y = 2x2–4

(i) Completeeachtableanddrawthecorrespondinggraph.Foreachgraphwritedown

(a)theequationoftheaxisofsymmetryofthegraph, (b)thecoordinatesoftheturningpoint, (c)themaximumorminimumvalueofthefunction.

3.Foreachofthefunctionsgivenin(a)to(d)below,prepareatableofvaluestodrawthegraphofthefunction,consideringtheintegralvaluesofxintherange–3<x<3.

Foreachfunction, (i)drawthecorrespondinggraph (ii)writedowntheequationoftheaxisofsymmetryofthegraph (iii) indicate the turning point on the graph and write downwhether it is a

maximumoraminimum(iv)writedownthemaximumorminimumvalueofthefunction. (a)y = x2+4 (b)y = 4–x2 (c)y =–(2x2+3) (d)y =4x2–5

37For free distribution

21.6 Finding the interval of values of x corresponding to an interval of values of y, for a function of the form y = ax2 + b

Letusidentifyhowtheintervalofvaluesofx correspondingtoanintervalofvaluesofyisfoundforafunctionwithaminimumvalue,byconsideringthegraphofthefunction y = x2–3.Letusfindtheintervalofvaluesofxforwhichthevalueofthefunctionislessthan6;thatis y < 6' Letusfirstdrawthegraphofy=x2–3.

y=x2–3 x –4 –3 –2 –1 0 1 2 3 4 x2 16 9 4 1 0 1 4 9 16–3 –3 –3 –3 –3 –3 –3 –3 –3 –3 y 13 6 1 –2 –3 –2 1 6 13

^–4" 13& ^–3" 6& ^–2" 1& ^–1" –2& ^0" –3& ^1" –2& ^2" 1& ^3" 6& ^4" 13&

–3–4 –2 –1 1 2

20

6

10

4

8

12

3 4–2

–4

x

y14

y = 6

y=x2–

3

Toidentifytheregionofthegraph,belongingtotheintervalofvalues y < 6,letusdrawthestraightliney =6.Intheregionofthegraphbelowtheliney =6,they coordinatetakesvalueslessthan6.Thisregionofthegraphhasbeenindicatedbyadarkline.Letusdrawtwolinesparalleltothey-axis,fromthepointsofintersectionofthegraphandtheliney =6,uptothex -axis.Letusmarkthetwopointsatwhichthesetwolinesmeetthex-axis(–3and+3).Theintervalofvaluesofxbetweenthesetwopointsistheintervalofvaluesofxforwhich y < 6' Thatis,whenthevalueofxisgreaterthan−3andlessthan+3,thevalueof y < 6' Therefore,theintervalofvaluesofxforwhichthefunctiony = x2–3satisfiesthecondition y < 6 is –3 < x < 3'

38 For free distribution

Example 1

Forthefunctiony = x2–4,(i)findthevaluesofxforwhich '

(ii)whatistheintervalofvaluesofxforwhichthefunctionisincreasingpositively?(iii)whatistheintervalofvaluesofxforwhichthefunctionisdecreasingpositively?(iv)whatistheintervalofvaluesofxforwhichthefunctionisincreasingnegatively?(v)whatistheintervalofvaluesofx forwhichthefunctionisdecreasingnegatively?Firstletusdrawthegraph. y = x2–4

x −4 −3 −2 −1 0 1 2 3 4x2 16 9 4 1 0 1 4 9 16−4 −4 −4 −4 −4 −4 −4 −4 −4 −4y 12 5 0 −3 −4 −3 0 5 12

^–4" 12& ^–3" 5& ^–2" 0& ^–1" –3& ^0" –4& ^1" –3& ^2" 0& ^3" 5& ^4" 12&

–3–4 –2 –1 1 2

2

0

6

10

4

8

12

3 4–2

–4

x

y14

y=x2–

4

(i)Theportionofthegraphforwhich ,istheportiononandabovethestraightliney =0.Thecorrespondingvaluesofxarethosewhicharelessthanorequalto–2orgreaterthanorequalto+2.Thatis, x ≤ –2 or x ≥ 2

(ii) x > 2.(iii) x < –2.(iv)0< x < 2.(v)–2< x < 0.

39For free distribution

Exercise 21.6

1.Drawthegraphofy=3–2x2andfindtheintervalofvaluesofxforwhichy>1.

2.Drawthegraphofy=2x2–4andfind,(i)theintervalofvaluesofxforwhichy<–3.(ii)theintervalofvaluesofxforwhichthefunctionincreasesnegatively.(iii)theintervalofvaluesofxforwhichthefunctionincreasespositively.(iii)theintervalofvaluesofxforwhichthefunctiondecreasespositively.(iv)theintervalofvaluesofxforwhichthefunctiondecreasesnegatively.

21.7 Finding the roots of an equation of the form ax2 + b = 0 by considering the graph of a function of the form y = ax2 + b

Letusconsiderforexamplehowtherootsoftheequationx2–4 =0arefound.Todothis,weneedtofirstdrawthegraphofthefunctiony=x2–4.

y=x2–4x –3 –2 –1 0 1 2 3

x2 9 4 1 0 1 4 9

–4 –4 –4 –4 –4 –4 –4 –4y 5 0 –3 –4 –3 0 5

^–3" 5& ^–2" 0& ^–1" –3& ^0" –4& ^1" –3& ^2" 0& ^3" 5&

–2 –1 1 2

–4

–3

0

2

–2

1

4

3

3 4–1

x

y5

–3–4^-–2" 0& ^2" 0& y=0

y=

x2 –4

40 For free distribution

Thetwopointsatwhichthegraphofthefunction y = x2–4 cutsthe x - axisarex = –2 and x = +2. Thatis,when x = –2 and x = +2,the y coordinateofthegraphis 0. Therefore,when x = –2 and x = +2, wehave x2–4 = 0. Thatis, x = –2 and x = +2 satisfytheequationx2–4 = 0. Toputthisinanotherway,therootsoftheequation x2–4= 0 are 2 and –2.

Exercise 21.7

1.Complete the following table of values to draw the graph of the functiony = 9 – 4x2.

x –2 –1 – 0 + 1 2

y –7 5 8 9 ___ 5 –7

(i) Usingthetable,drawthegraphofthefunctiony = 9 – 4x2

(ii) Usingthegraph,findtherootsoftheequation9 – 4x2 =0

2.Prepare a table of valueswith –3 < x < 3 to draw the graph of the functiony = x2 –1'

(i) Drawthegraphofy = x2 –1' (ii) Usingthegraph,findtherootsofx2 –1 = 0'

3.Prepare a table of valueswith –3 < x < 3 to draw the graph of the functiony = 4 – x2'

(i) Drawthegraphof y = 4 – x2'

(ii) Usingthegraph,findtherootsof 4 –x2 =0'

4.Prepareasuitabletableofvaluesanddrawthegraphof y = x2–9. (i) Drawthegraphof y = x2–9. (ii) Usingthegraph,findtherootsof9–x2 =0'

41For free distribution

21.8 Verticle displacement of graphs of functions of the form y = ax2 + b

Considerthegraphsgivenbelowwhichyouhavestudiedearlier.

y=x2 +3

3 4

–2x–3–4 –2 –1 1 2

2

0

6

10

4

8

12

y

y=x2

y=x2–2

•Observethat,by translating the graph of y = x2 by 3 units vertically upwards, the graphcorrespondingtothefunctionwiththeequationy = x2+3isobtainedandalsoby translating thegraphofy = x2 by2unitsverticallydownwards, thegraphcorrespondingtothefunctionwiththeequationy = x2–2isobtained.

Observethefollowingtable.

Equationofthegraph Minimumpoint Axisofsymmetryy = x2

y = x2+3y = x2–2

^0" 0&

^0" 3&

^0" –2&

x=0x=0x=0

Accordingly,² ifwetranslatethegraphofy = x2by6unitsverticallyupwards,theequationofthegraphofthecorrespondingfunctionwillbey = x2+6'

² ifwetranslatethegraphofy = x2by4unitsverticallydownwards,theequationofthecorrespondingfunctionofthegraphwillbey = x2–4'

² ingeneral,theequationofthegraphobtainedbytranslatingthegraphofafunctionoftheformy = ax2+b verticallyupwardsordownwardsbycunitsisrespectivelyy = ax2+b + c ory = ax2+b –c.

42 For free distribution

Exercise 21.81.Ifthegraphofthefunctiony = x2+2"(i)movesupwardsalongtheyaxisby2units(ii)movesdownwardsalongtheyaxisby2unitswritetheequationofthegraph.

2.Ifthegraphofthefunctiony = –x2"(i)movesupwardsalongtheyaxisby3units(ii)movesdownwardsalongtheyaxisby3unitswritetheequationofthegraph.

3.Ifthegraphofthefunctiony = 2x2+5"(i)movesupwardsalongtheyaxisby6units(ii)movesdownwardsalongtheyaxisby6unitswritetheequationofthegraph.

Miscellaneous Exercise

1. Thecoordinatesoftwopointsonastraightlinegraphare(0,3)and(3,1).(i)Calculatethegradientofthegraph.(ii)Findtheinterceptofthegraph.(iii)Writedownthefunctionofthegraph.

2.Without drawing the graph, providing reasons show that the points (–1, –3),(2,4)and(4,6)lineonthesamestraightlinegraph.

3.Without drawing the graph, providing reasons show that the points (–2, –8),(0,–2),(3,7)and(2,4)lieonthesamestraightlinegraph.

4.Drawthegraphofthefunction . (i)Usingthegraph,findtheintervalofvaluesofxforwhich '

(ii)Usingthegraph,findtheintervalofvaluesofxforwhichthevalueofthe

functionislessthan–1.

5. Construct the table of values to draw the graph of the function y = 3 – 2x2 intheinterval–2 ≤ x ≤ 2'

(i)Usingthetableofvalues,drawthegraphofy = 3 –2x2.(ii)Usingthegraph,findtherootsoftheequation 3 –2x2=0' (iii)Writedown theequationof thegraphswhich isobtainedwhen theabove

graphisshiftedupwardsby2units.