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IM2Sem1FinalReviewG 1
IM2Semester1FinalExamReviewG(StudyGuideQuestions64-72)
ConvertingEquationsandGraphingQuadratics
Torewriteaquadraticfrommodifiedvertexformintostandardform(problems64–66):
Startbyrewritingthe 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔 !as(𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔)(𝑖𝑡𝑠𝑒𝑙𝑓). Leavethenumberattheend(k)whereitis.Youwilladdorsubtractitlast.Next,distributeeverypartofthefirstparentheseswitheverypartofthesecondparentheses. Ifyoudidthisstepcorrectly,youwillhave4terms(andoneextra(the“k”)hangingoutintheback).Lastly,combineliketermstowritetheequationintheform𝑓 𝑥 = 𝑎𝑥! + 𝑏𝑥 + 𝑐. Addorsubtracttermswiththesamevariables(𝑥! & 𝑥!, 𝑥 & 𝑥, 𝑛𝑢𝑚𝑏𝑒𝑟 & 𝑛𝑢𝑚𝑏𝑒𝑟).
1.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 7𝑥 + 4 ! + 1
2.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 2𝑥 − 8 ! + 3
3.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 9𝑥 − 8 ! − 5
4.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 6𝑥 − 5 ! − 2
5.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 5𝑥 + 1 ! − 4
6.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 8𝑥 + 2 ! + 3
7.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 2𝑥 − 9 ! − 3
8.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 6𝑥 − 1 ! − 2
9.Rewritethequadraticequationinstandardform.
𝑓 𝑥 = 3𝑥 + 8 ! + 9
Tocompletethesquaretofindpandq(problems67–69):
Therearetwowaystosolvethisproblem.OPTION1:Determinethevertextowritetheprobleminvertexform. ℎ = − !
!!,thenplugℎinforallthe𝑥-valuesintheoriginalequationtofind𝑘.
Rewritetheproblemintheform𝑎 𝑥 − ℎ ! + 𝑘 = 0Then,movektotheothersidebyaddingorsubtractingit.Identify𝑝&𝑞basedoffoftheequation 𝑥 − 𝑝 ! = 𝑞 Basically,𝑝 = ℎ,and𝑞 = −𝑘(switchk’ssign).OPTION2(see#19-27onReviewF):Startbymoving𝑐totheothersideoftheequalsign.
Then,rewrite𝑥! + 𝑏𝑥 = −𝑐as 𝑥 + !!
!= −𝑐 + !
!
!
Dividethenumberinfrontofthex-termby2.Writethatnumberinsquaredparenthesesontheleft:
𝑥 𝑠𝑎𝑚𝑒 𝑠𝑖𝑔𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 2 !=
Whatevernumberyouputontheleft,squareitandaddthatnumbertothenumberontheright.Thensimplifytherightside&identify𝑝&𝑞basedoffoftheequation 𝑥 − 𝑝 ! = 𝑞.
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IM2Sem1FinalReviewG 2
Forexample:𝑥! − 100𝑥 − 8 = 0 𝑥! − 100𝑥 = 8100 ÷ 2 = 50 & 50 ! = 2500
𝑥 − 50!= 8 + 2500
𝑥 − 50 ! = 2508 𝑝 = 50, 𝑞 = 2508 10.Theequation𝑥! + 10𝑥 + 1 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
11.Theequation𝑥! − 6𝑥 − 24 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
12.Theequation𝑥! + 18𝑥 − 5 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
13.Theequation𝑥! + 6𝑥 + 15 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
14.Theequation𝑥! − 8𝑥 + 23 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
15.Theequation𝑥! + 34𝑥 + 2 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
16.Theequation𝑥! + 20𝑥 − 13 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
17.Theequation𝑥! − 22𝑥 − 17 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
18.Theequation𝑥! − 12𝑥 + 4 = 0canbetransformedintoanequationoftheform 𝑥 − 𝑝 ! = 𝑞,where𝑝and𝑞arerationalnumbers.Completethetablebelowwiththevaluesof𝑝and𝑞.
Constant Value𝑝 𝑞
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IM2Sem1FinalReviewG 3
Tographaquadraticinvertexform(problems70-72):
Startbydeterminingyourvertex(ℎ, 𝑘),whichisgiventoyouinvertexform:𝑓 𝑥 = 𝑎 𝑥 − ℎ ! + 𝑘.Graphthatpoint.Youneedaminimumof5points.Youmusthavethevertexandcrossortouchboththex-andthey-axes. Theonlyreasonsforyourgraphnottocrossbothaxeswouldbe
1. Iftheprovidedgraphingspacewasnotbigenoughtoallowyoutocrossthem.2. Ifthegraphcannevercrossaxis(therootsareimaginary).
Thereareseveralwaystofindtheotherpointsthatyouneed.Theprocessbelowusesanxytable.x y
Putyourvertexinthemiddleofanx-ytable
Onthexside,includeasmanyx’sinbothdirectionsasyouthinkyouneed(movebyoneeachtime).
Now,startwithoneofthex’sthatis1awayfromthevertex.
Plugitintotheoriginalequation.Simplifytogety.Putthaty-valueinthetable
©ittothematchingxontheothersideofthevertex.Graphthosetwopoints.
Repeatwiththex’sthatare2away,andsoonuntilyouhavetouched
orcrossedeachaxis.
Ifyourgraphlooksstrange,youhavelikelymadeamistake–gobackandcheckyourwork.
3less
Same
Same
2less
Same
1lessthanthevertexx
Vertexx Vertexy
1morethanthevertexx
2more
3more
19.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = − 𝑥 − 3 ! + 8
20.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = − 𝑥 + 4 ! + 10
21.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = 𝑥 + 1 ! − 2
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IM2Sem1FinalReviewG 4
22.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = 2 𝑥 − 1 ! − 8
23.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = − 𝑥 − 4 ! + 3
24.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = 𝑥 + 2 ! − 4
25.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = − 𝑥 − 2 ! − 3
26.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = 𝑥! − 1
27.Graph.Labelthevertexandaxisofsymmetry.
𝑓 𝑥 = − 𝑥 + 3 ! + 5
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IM2Sem1FinalReviewG 5
IM2Semester1FinalExamReviewGAnswers
1.𝑓 𝑥 = 49𝑥! + 56𝑥 + 17 2.𝑓 𝑥 = 4𝑥! − 32𝑥 + 67 3.𝑓 𝑥 = 81𝑥! − 144𝑥 + 594.𝑓 𝑥 = 36𝑥! − 60𝑥 + 23 5.𝑓 𝑥 = 25𝑥! + 10𝑥 − 3 6.𝑓 𝑥 = 64𝑥! + 32𝑥 + 77.𝑓 𝑥 = 4𝑥! − 36𝑥 + 78 8.𝑓 𝑥 = 36𝑥! − 12𝑥 − 1 9.𝑓 𝑥 = 9𝑥! + 48𝑥 + 7310.𝑝 = −5; 𝑞 = 24 11.𝑝 = 3; 𝑞 = 33 12.𝑝 = −9; 𝑞 = 8613.𝑝 = −3; 𝑞 = −6 14.𝑝 = 4; 𝑞 = −7 15.𝑝 = −17; 𝑞 = 28716.𝑝 = −10; 𝑞 = 113 17.𝑝 = 11; 𝑞 = 138 18.𝑝 = 6; 𝑞 = −3219.
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