00020002010015QuadraticEquation.ACT0002020015EquationWithRoots.EAC01000000b08dDkx[Quadratic Equations[With Their Roots\Author\PMun Chou, Fong Mathsroof Consultancy Malaysia. e-mail: mcfong@mathsroof.com[[ OBJECTIVE[I. To construct quadratic[equations with its roots;[[II. To learn that many[quadratic equations have[the same roots and then[discover their general[form.[[ Example 1:[The roots of a quadratic[equation are given as -3 [and 2.4. The task is to[construct the suitable[quadratic equation with [these given roots.[[ Solution:[One way is through the[expansion of (-a)(-b)[where a and b are the [roots of the equation.[Tap Expansion strip to [ see how...[\Expansion StriplR clear_a_zRdoneR(-(-3))(-2.4)Rx+3x-125Rexpand(x+3x-125)Rx2+3x5-365RGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O[With given real roots of[-3 and 2.4, we obtained[the quadratic equation as[x2+3x5-365=0.[[Exploration 1:[An interesting question to[ask now: Is the equation[obtained an unique one?[[To answer the question,[lets draw the graph of[y1=x2+3x5-365 and its[ variations.[- Tap on GraphStrip 1.[- Draw y1 and observe [ that the curve passes [ through -3 and 2.4 [ on x-axis.[- Now select function y2[ and y3. Draw all three[ y1, y2 and y3 on the[ same axes. [- Tap [Analysis][G-Solve][  [Root] to find the [ roots of all three y1,[ y2 and y3. [\ GraphStrip 1Graph2D Graph3D LISTSYS4NModify $ STATCALC NSTATSYS \NSequencel, Sheet| Sheet3D| SolveEqSolveLwr SolveUpr StupFLG1(StupListD StupPict$ ViewWind< summaryXl y1$Hy2(Hy3$H4 @ L X d p |            ! " # $$ %0 &< 'H (T )` *l +x , - . / 0L 1X 2d 3p 4| 5 E F H I J K L M N O P Q R$ S0 T< ]H ^L _P `T aX b\ ` l x   system]listsystem^]=system_^system`_systema`systemb~ aR6,-4.477bT system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $  $(4<HLX\hp|$(4<HP\`lxxTx-7.7*x/p``-3/10x/Yp7.7pf(x)x-14.8x/H`-Tx0R+Tx16`f(x)x2R+Tx2Q+Tx2Tf(x))x47.47x/tpx-729/100-36)`x56.7100g 3xx^2+3*x/5-36/5 3x2*(x^2+3*x/5-36/5) 3x-(x^2+3*x/5-36/5)`P`P`p`p uRcx`dG6"(10qy`#YuY 000000 $0<HT`l```` `BRcy T[[(1) Clearly the three[curves pass x=-3 and[x=2.4. What does this[imply?[[(2) State 2 relationships[between y2, y3 and y1.[[Now try the next activity[before answering (3),(4)[and (5).[ GRAPH MODIFY[- Tap on GraphStrip 2.[- Select y1 and graph it[- Tap on [Analysis] [ [Modify].[When equation of the[curve appears in the[graph message box,[select only the 'A' at[front but not the whole[ equation.[- Now tap on the right[ and left graph arrow[ of to modify A.[\ GraphStrip 2lGraph2Dh Graph3D| LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceD, Sheetp| Sheet3D| SolveEqhSolveLwrl SolveUprx StupFLG1(StupListD StupPict$ ViewWind y10(Hy2X$Hy3|$H           $ 0 < H T  ` !l "x # $ % & ' ( ) * + , - . 0 1 2, 38 4D 5P E\ Fh Ht I J K L M N O P Q R S T ] ^ _ ` a  b$ ( 4 @ L X system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $  3xA*(x^2+3*x/5-36/5) 3xx^2+3*x/5-36/5-A 3xA*x^2+3*x/5-36/5`P`P`p`p `(10qy`#YuY Y)`` R[Watch as the curve is[transformed according to[ values of A. [[Repeat GRAPH MODIFY but[first with y2 then follow[by y3.[[(3) Describe what you[may have observed in[the exercise of modifying[A in y1, y2 and y3.[[(4) State 2 examples of[quadratic equations which[have roots of -3 and[2.4.[[(5) Which equation below[best describes quadratic[equations with roots of[ -3 and 2.4?[a) x2+3x5-365=A, [b) Ax2+3x5-365=0, or[c) Ax2+3x5-365=0[[###Explore 1 Notes###[(1)The graphs imply that[these equations have -3[and 2.4 as their roots,[ and that x2+3x5-365=0[is not an unique answer.[[(2)The relationships are:[(i)y2=2y1 and (ii)y3=-y1 [[(3) One would observe[that the roots of the[equation remain the same[as -3 and 2.4 while A of[y1 is being modified.[ Modify A of y2 shifts[the curve vertically while[modifying A of y3 shrinks[ the curve.[[(4),(5)[From our modify exercise[in (3), general form of[ quadratic equations with[ roots of -3 and 2.4 is[ obviously[ A(x2+3x5-365)=0,[where A is real,[#######################[[ Exercise 1:[Use similar approach to[show that many quadratic[equations may also be[formed when the roots[are repeated and real, [such as -2 and -2.[\Use here|R clear_a_zRdoneR(-(-2))(-(-2))Rx+22RGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O\Use hereGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O[[The approach we used in[Exploration 1 is not very[effective if the equations[we are discussing have[imaginary roots.[[Exploration 2:[To understand why, we[graph y1=x2+4x+5 and[its variations of y2 and[y3 in GraphStrip 3 to[ illustrate.[[###Explore 2 Notes###[ - Note that x2 +4x+5=0[has imaginary roots,since[its discriminant is [ 42 -415=-4[which is less than 0.[#######################[\ GraphStrip 3Graph2D| Graph3D LISTSYS4NModify $ STATCALC NSTATSYS \NSequenceX, Sheet| Sheet3D| SolveEq|SolveLwr SolveUpr StupFLG1(StupListD StupPict$ ViewWind( y1DHy2` Hy3$H           ( 4 @ L X  d !p "| # $ % & ' ( ) * + , - . /D 0\ 1h 2t 3 4 5 E F H I J K L M N O P Q( R4 S@ TL ]X ^\ _` `d ah bl p |    system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $   3xx^2+4*x+5 3x2*(x^2+4*x+5) 3x-3*(x^2+4*x+5)`P`P``88 `(10qy`#YuY  $0<HT`lx`p@`` `p@`P`P``&1WG7 S[The graphs show that [these three curves do[not intersect with x-axis[Hence we would not be[able to graphically show[they meet at some points[ on x-axis. [[We can however employ[this strategy:[Show that the solution[of B(x2+4x+ 5)=0 always[produce the same exact[solution for many non-[zero real values of B.[ [- Tap on Explore Strip.[- An initial value of 1 is[ assigned to B.[- Try assigning different[ real values to B.[ Remember to tap [EXE][ after each change.[\Explore Strip  hR clear_a_zRdoneR1 BR1RB(^2+4+5)=0eq1Rx2+4x+5=0R solve(eq1)Rx=-2-,x=-2+RGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O[Note that for any real[values of B assigned,[the roots are always[-2- and -2+ .(except[when 0 is assigned to B)[[This result is suffice to[imply that infinitely many[ quadratic equations have[imaginary roots -2- and[-2+.[ \About   is a symbol denoting P(-1). It is important in the discussion on Complex Number. The general form of complex number is a+b, where a, b are real. [[eActA UxB Uxeq Eeq1LLF 0 ;xxeq244F 1  xeq3((F 1 xeq5S)/(02000eFindMaxMin.EAC01000000c2a5y=-3x^(2)+2x+1`gE(0 `gE(0 vrALCuxy@  -3x^2+2x+1Dkx[Finding Max And Min[ With ROOTS\Author\PMun Chou, Fong Mathsroof Consultancy Malaysia. e-mail: mcfong@mathsroof.com[[ OBJECTIVE[To algebraically determine[the maximum or minimum[point of quadratic curves[using roots of the[corresponding quadratic[ equation.[[Exploration 1:[We begin by working on[this exercise to discover[a relationship between[the  curve of A2+b+c[and  coefficient of 2,[which is A. Now[- Tap on Geometry Strip.[- Replace the coefficient[ of 2 in Geometry Link[ with -3,,2,-1 and 4.[ Remember to tap [EXE][ after each replacement[]\Geometry Strip[[Try describe what you[observe, then tap to see[notes...\Exploration Notes We may conclude from this exercise that: - If coefficient of is positive, the curve of a+b+c is a minimum curve/has minimum point, On the other hand, - if coefficient of is negative, then the curve is a maximum curve/has maximum point. [[ Example 1:[The task here is to find[the minimum or maximum[point of y=2+6+7.[Let this point be (x,y)[The coefficient of 2 is[1, which is > 0.[From Exploration 1, we[can conclude that (x,y)[is a minimum point.[[Apparently x is the[axis of symmetry for[the curve of 2+6+7,[and it can be  determined[with  sum of roots2.[y is found by substitution[of x into the  function of[y=2+6+7.[ [We begin the solution by[first locating the roots[ graphically.[- Tap on GraphStrip.[ - Select 2+6+7 and[ drag it into the window.[\ GraphStrip Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ABV#r`` O[Once the curve is drawn,[tap on [Analysis][G-Solve][ [Root] to see that the[roots are -4.414213562372[and -1.585786437627[[But in finding (x,y) we[prefer the roots in exact[form. A good technique [to get  the exact roots is[by using quadratic formulae.[Recall our equation is[ 2+6+7=0[ Tap to see...\ Use Formulae  R clear_a_zRdoneR1aR1R6bR6R7cR7R-b+b2-4ac2aRoot1R- 1.585786438R-b-b2-4ac2aRoot2R- 4.414213562Rsimplify(Root1)R-3+2Rsimplify(Root2)R-3-12RGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O[[The last 2 operations[show the exact roots as[-3+2 and -3-12.[Therefore x is, R(-3+2)+(-3-12)2R-3[[ While y is x2+6x+7, orR(-3)2+6(-3)+7R-2[[So the minimum point of[y=2+6+7, or (x,y),[ is (-3,-2).[[ Exercise 1:[Use the method discussed[to determine the maximum[ or minimum  point of [ y=-22+5-1.\Use here Graph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY (v@d` O\Use here R clear_a_zRdoneR1aR1R1bR2R1cR1RGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O[[Exploration 2:[Lets extend the method[discussed in Example 1[to equation with imaginary[roots. For example the [ equation 52 -3+9=0.[Our first goal is to find[the max/min point. Lets[call this point (x,y). [[As its coefficient of 2[is >0, we conclude that [(x,y) is a minimum point.[[Now tap on CALC strip[and assign appropriate[ values to A,  B and C to[solve 52 -3+9=0.[\CALCR clear_a_zRdoneR1AR1R1BR1R1CR1R-B+B2-4AC2ARoot1R-0.5+ 0.8660254038R-B-B2-4AC2ARoot2R-0.5- 0.8660254038Rsimplify(Root1)R-12+123Rsimplify(Root2)R-121+3RGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! " # $ % & ' ( ) *, +8 ,D -P .\ 0h 1t 2 3 4 5 E F H I J K L M N O P( Q4 R@ SL TX ]d ^h _l `p at bx |     system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $ `P`P`p`p `(10qy`#YuY ` O[What are our Root1 and[Root2?[Put Root1 and Root2 in[place of , below to[determin e (x,y).[R()+()2xR0.5P+QR5x2-3x+9R1.25P+Q2-1.5P+Q+9[[At this juncture we have[found (x,y). As a final[step we want to verify[graphically that (x,y)[is indeed the minimum of[52-3+9. [ [- Tap on GraphStrip 2.[- Draw y1=52-3+9.[- Use [Analysis][G-Solve][ [Min] to locate the[ minimum point.[Does this point consistent[with our (x,y)?[ \ GraphStrip 2Graph2D@ Graph3DT LISTSYS`4NModify $ STATCALC NSTATSYS \NSequence, SheetH| Sheet3D| SolveEq@SolveLwrD SolveUprP StupFLG1\(StupListD StupPict$ ViewWind y1 H( 4 @ L X d p |           ! " # $ %$ &0 '< (H )T *` +l ,x - . 0 1 2 3 4 5 E F H I J K L, M8 ND OP P\ Qh Rt S T ] ^ _ ` a b      system]listsystem^]=system_^system`_systema`systemb~ aseq_histbNewFolde system]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=systemsystem]listsystem^]=system  0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D 0@PSheet1^Sheet2^Sheet3^Sheet4^Sheet5^SheetSheet3D i Dkx $  3x5*x^2-3*x+9`P`P``)) `(10qy`#YuY `Q``&1WG7 P[[ Tap to see...\Exploration NotesThe minimum point (x,y) is (0.3,8.55). Maximum or minimum of any quadratic curve can be located with this ROOT method regardless of the condition of its roots. [[ Exercise 2:[Find the maximum/minimum[of y=-57x2+8x-1152.[Having found the roots,[verify them by graphing[ the function.[\ Calculate4R clear_a_zRdoneRGraph2D, Graph3D@ LISTSYSL4NModify $ STATCALC NSTATSYS \NSequence, Sheet4| Sheet3D| SolveEq,SolveLwr0 SolveUpr< StupFLG1H(StupListpD StupPict$ ViewWind     $ 0 < H T ` l x       ! 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