21 Graphs - Minister of Education
Transcript of 21 Graphs - Minister of Education
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
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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
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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
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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
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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)
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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).
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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.
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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.
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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.
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•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&
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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.
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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.
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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.
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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.
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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.