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MultipleTransformationsforAbsoluteValueandQuadraticFunctionsWhenfindingtheequationofabsolutevalueorquadraticfunctionsfromagraphintheform𝑓(𝑥) = 𝑎 𝑥 − ℎ ! + 𝑘or𝑓(𝑥) = 𝑎 𝑥 −ℎ + 𝑘,followthesesteps:
1.) Figureoutwhatkindofparentfunctionitis:• V-shapedàAbsolutevaluefunctionso𝑓(𝑥) = 𝑎 𝑥 − ℎ + 𝑘• U-shaped/parabolaà𝑓(𝑥) = 𝑎 𝑥 − ℎ ! + 𝑘
2.) Findthevertex.Thiswillgiveyouhandk.3.) Plugthevertexintotheaboveequationforthecorrectparentfunction.Remember,ifhisnegative,itwillbecome+insidethe
absolutevalue/parenthesessincetwonegativesequalsapositive.4.) Ifthefunctionisopeningdownward,youknowit’sareflectionandtherewillbeanegativesigninfrontoftheabsolute
value/parentheses.5.) Lastly,finda.Todothis,findanotherpointthat’sonyourgraphbesidesthevertex.Ifyouusethevertex,thiswillnotwork!
Plugthepointinforxandy(𝑓(𝑥))inyourequation.Youshouldhavethehandkalreadyfilledinfromthevertexandyounowwillhavexandyfilledinaswell.Theonlyvariableleftshouldbea!Solveyourequationfora.
6.) Inyourfinalequation,youshouldhavehandkfromthevertexandafromthepreviousstepfilledin.Youshouldnothaveanythingfilledinforxandyasthispointisdependentontheactualgraph.Voila!You’redone!
Example#1:
Step1:Sincethisisu-shaped/parabola,usethegeneralformofthefunction:𝑓(𝑥) = 𝑎 𝑥 − ℎ ! + 𝑘Step2:Findthevertexà(2,1).Thus,h=2andk=1.Step3:Plughandkintotheequation:𝑓(𝑥) = 𝑎 𝑥 − 2 ! + 1Step4:Sincetheparabolaisopeningup,itisnotareflectionandthus,therewillbenonegativesign.Step5:Weneedtopickanotherpointontheparabolathat’snotthevertex.Forthis,I’lluse(5,4).Now,plugthisintoyourequationforstep3.x=5andyor𝑓(𝑥) =4.Soournewequationis4 = 𝑎 5 − 2 ! + 1.Solveyournewequation.
4 = 𝑎 5 − 2 ! + 14 = 𝑎 3 ! + 14 = 9𝑎 + 13 = 9𝑎
𝑎 =13
Step6:Plugthevaluesforh,k,andabackintoyourgeneralformoftheequationandyou’redone!
𝑓(𝑥) =13𝑥 − 2 ! + 1
Example#2:
Step1:Sincethisisv-shaped,usethegeneralformofthefunction:𝑓(𝑥) = 𝑎 𝑥 − ℎ + 𝑘Step2:Findthevertexà(-1,2).Thus,h=-1andk=2.Step3:Plughandkintotheequation:𝑓(𝑥) = 𝑎 𝑥 + 1 + 2**Note–Sincehisnegative,itbecomes+insidetheabsolutevalue.Step4:Sincetheparabolaisopeningdown,itisareflectionandthus,therewillbeanegativesigninfrontoftheabsolutevalue.Step5:Weneedtopickanotherpointontheparabolathat’snotthevertex.Forthis,I’lluse(0,0).Now,plugthisintoyourequationforstep3.x=0andyor𝑓(𝑥) =0.Soournewequationis0 = 𝑎 0 + 1 + 2.Solveyournewequation.
0 = 𝑎 0 + 1 + 20 = 𝑎 1 + 20 = 1𝑎 + 2−2 = 1𝑎𝑎 = −2
Step6:Plugthevaluesforh,k,andabackintoyourgeneralformoftheequationandyou’redone!𝑓 𝑥 = −2 𝑥 + 1 + 2