Multiple Transformations for Absolute Value and …...3.) Plug the vertex into the above equation...
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Multiple Transformations for Absolute Value and Quadratic Functions When finding the equation of absolute value or quadratic functions from a graph in the form () = − ℎ ! + or () = − ℎ + , follow these steps: 1.) Figure out what kind of parent function it is: • V-shaped à Absolute value function so () = − ℎ + • U-shaped/parabola à () = − ℎ ! + 2.) Find the vertex. This will give you h and k. 3.) Plug the vertex into the above equation for the correct parent function. Remember, if h is negative, it will become + inside the absolute value/parentheses since two negatives equals a positive. 4.) If the function is opening downward, you know it’s a reflection and there will be a negative sign in front of the absolute value/parentheses. 5.) Lastly, find a. To do this, find another point that’s on your graph besides the vertex. If you use the vertex, this will not work! Plug the point in for x and y(()) in your equation. You should have the h and k already filled in from the vertex and you now will have x and y filled in as well. The only variable left should be a! Solve your equation for a. 6.) In your final equation, you should have h and k from the vertex and a from the previous step filled in. You should not have anything filled in for x and y as this point is dependent on the actual graph. Voila! You’re done! Example #1: Step 1: Since this is u-shaped/parabola, use the general form of the function: () = − ℎ ! + Step 2: Find the vertex à (2,1). Thus, h=2 and k=1. Step 3: Plug h and k into the equation: () = − 2 ! + 1 Step 4: Since the parabola is opening up, it is not a reflection and thus, there will be no negative sign. Step 5: We need to pick another point on the parabola that’s not the vertex. For this, I’ll use (5,4). Now, plug this into your equation for step 3. x=5 and y or () = 4. So our new equation is 4 = 5 − 2 ! + 1 . Solve your new equation. 4 = 5 − 2 ! + 1 4 = 3 ! + 1 4 = 9 + 1 3 = 9 = 1 3 Step 6: Plug the values for h, k, and a back into your general form of the equation and you’re done! () = 1 3 − 2 ! + 1 Example #2: Step 1: Since this is v-shaped, use the general form of the function: () = − ℎ + Step 2: Find the vertex à (-1,2). Thus, h=-1 and k=2. Step 3: Plug h and k into the equation: () = + 1 + 2 **Note – Since h is negative, it becomes + inside the absolute value. Step 4: Since the parabola is opening down, it is a reflection and thus, there will be a negative sign in front of the absolute value. Step 5: We need to pick another point on the parabola that’s not the vertex. For this, I’ll use (0,0). Now, plug this into your equation for step 3. x=0 and y or () = 0. So our new equation is 0 = 0 + 1 + 2 . Solve your new equation. 0 = 0 + 1 + 2 0 = 1 + 2 0 = 1 + 2 −2 = 1 = −2 Step 6: Plug the values for h, k, and a back into your general form of the equation and you’re done! = −2 + 1 + 2
Transcript of Multiple Transformations for Absolute Value and …...3.) Plug the vertex into the above equation...
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
