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Question 22834: I believe this would be the best place to ask...Okay, this is a relation (set of ordered pairs) that I have to write an equation for. The set is :
(-3,10)(-2,5)(-1,2)(0,1)(1,2)(2,5)(3,10)
x : -3, -2, -1, 0, 1, 2, 3
y : 10, 5, 2, 1, 2, 5, 10
I'm supposed to divide the range differences by the domain differences :
range difference : 5, 3 1
domain difference : 1
But, if I divide 5/1, 3/1 and 1/1, I end up with 5, 3 and 1, as any number divided by 1 is that number. So where do I go from here and how do I get there???
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I believe this would be the best place to ask...Okay, this is a relation (set of ordered pairs) that I have to write an equation for. The set is :
SEE MY COMMENTS BELOW
(-3,10)(-2,5)(-1,2)(0,1)(1,2)(2,5)(3,10)…OK
x : -3, -2, -1, 0, 1, 2, 3………GOOD
y : 10, 5, 2, 1, 2, 5, 10 ………GOOD
I'm supposed to divide the range differences by the domain differences :THIS IS HINT GIVEN TO HELP YOU..OK…
range difference : 5, 3 1…THIS IS NOT CORRECT.IT IS A GOOD PRACTICE TO FOLLOW AN ORDER OR CONVENTION.SUBTRACT PREVIOUS ENTRY FROM SUCCEEDING ENTRY ALWAYS.SO THE RANGE DIFFERENCE IS –5,-3,-1,1,3,5.
domain difference : 1 ..CONSTANT ALWAYS..GOOD.HERE YOU FOLLOWED THE CONVNTION I MENTIONED ABOVE
But, if I divide 5/1, 3/1 and 1/1, I end up with 5, 3 and 1, as any number divided by 1 is that number…OK ..BUT YOU GET –5,-3,-1,1,3,5 AS WE CORRECTED ABOVE.
So where do I go from here and how do I get there???…LOOK AT THE OBJECTIVE..WE HAVE TO FIND A PATTERN..A HINT IS GIVEN..WE USED THE HINT AND GOT THE NUMBERS AS –5,-3,-1,1,3,5…THERE IS A PATTERN HERE..THEY STARTED WITH –5 AND INCREASED BY A CONSTANT DIFFERENCE OF 2 ..THAT GIVES US THE CLUE TO DERIVE THE RELATION
LET US CALL THE FIRST NUMBER AS X AS YOU NAMED IT .SECOND NUMBER IS Y.
SO WE GOT (X1,Y1),(X2,Y2)…ETC
AND NOW YOU HAVE TO FIND A RELATION BETWEEN X AND Y BY ANALYSING THEM SEPERATELY OR TOGETHER BY COMBINING THEM
LET ME SHOW YOU BOTH METHODS.
FIRSTLY LET US ANALYSE SEPERATELY.WE IND THAT BOTH ARE LINKED TO N THE NUMBER OF THE TERM.SO LET US TRY TO RELATE EACH WITH N WHICH WILL THEN GIVE US WHAT IS CALLED AS PARAMETRIC RELATION.THAT IS X IS RELATED TO N AND Y IS RELATED TO N .SO INDIRECTLY X AND Y ARE RELATED THROUGH A PARAMETER N.
LET US TAKE UP X FIRST.WE FOUND
X2-X1=1
X3-X2=1
X4-X3=1
X5-X4=1
Etc…..
X(N+1)-X(N)=1
LET US ADD THEM ALL TOGETHER..WE FIND THAT X2,X3,X4…X(N) CANCEL OUT LEAVING ONLY X(N+1)-X1 ON THE L.H.S…ON THE R.H.S. WE GET 1+1+1…N TIMES =1*N=N
HENCE WE GET X(N+1)-X1=N
BUT X1=-3…SO ….X(N+1)=N+X1=N-3..WE CAN CHECK BCK BY PUTTING N=0,1,2,3,…..ETC…WE GET X(0+1)=X1= 0-3,….X(1+1)=X2=1-3=-2…ETC..OK.
NOW LET US SEE Y ,WE FOUND
Y2-Y1=-5
Y3-Y2=-3
Y4-Y3=-1
Y5-Y4=1
Y6-Y5=3
Y7-Y6=5
ETC…..
Y(N+1)-YN= -5 +(N-1)*2= -5+2N-2=2N-7…….LET ME EXPLIN THIS TO YOU.WE FIND THE NUMBERS ON THE R.H.S ARE –5,-3,-1 ETC….DIFFERING BY A CONSTANT VALUE 2..THIS IS CALLED ARITHMATIC PROGRESSION(A.P)……. 2 ND TERM I OBTAINED BY ADDING 2 O THE FIRT TERM .THIRD TERM IS GOT BY ADDING 2*2 TO THE FIRST TERM.FOURTH IS OBTAINED BY ADDING 2*3 TO THE FIRST TERM……N TH TERM IS OBTAINED BY ADDING (N-1)*2 TO THE FIRST TERM
THE FORMULAE ARE …NTH TERM =TN=A+(N-1)*D WHERE A IS THE FIRST TERM AND D IS THE CONSTANT DIFFERENCE.
SUM OF N TERMS OF AN A.P. IS GIVEN BY
SN=(N/2){2A+(N-1)*D}…WHICH WE SHALL USE BELOW
NOW AS WE DID EARLIER LET US ADD ALL THE ABOVE TO GET
Y(N+1)-Y1={-5+(-3)+(-1)+1+3………..+(2N-7)}=AS PER THE FORMULA ABOVE
Y(N+1)-Y1=(N/2){2*(-5)+(N-1)*2}=(N/2){-10+2N-2}=(N/2)(2N-12)
=(N/2)2(N-6)=N(N-6)=N^2-6N….BUT Y1=10 SO WE GET
Y(N+1)-Y1=Y(N+1)-10=N^2-6N….OR…….
Y(N+1)=N^2-6N+10……LET US CHECK BACK BY PUTTING N+0,1,2,ETC….
Y(0+1)=Y1=0-0+10=10…OK
Y(1+1)=Y2=1-6+10=5…..OK
Y(2+1)=Y3=2^2-6*2+10=2……OK
Y(3+1)=Y4=3^2-6*3+10=1…….OK
Y(4+1)=Y5=4^2-6*4+10=2………OK
Y(5+1)=Y6=5^2-6*5+10=5…….OK
Y(6+1)=Y7=6^2-6*6+10=10…OK…
SO NOW WE GOT …Y(N+1)=N^2-6N+10
THIS TOGETHER WITH X(N+1)=N-3….GIVES US THE RELATION BETWEEN X AND Y IN TERMS OF ACOMMON PARAMETER N.
NOW IF WE WANT X AND Y DIRECTLY RELATED WE HAVE TO ELIMINATE N ..WE NOTE THAT X(N+1)=N-3…OR N=X(N+1)+3…BY PUTTING THIS IN EQUATION FOR Y(N+1)=N^2-6N+10,WE GET A DIRET RELATION BETWEEN X AND Y …
Y(N+1)={X(N+1)+3}^2-6{X(N+1)+3}+10
IF YOU ARE MORE COMFORTABLE WITH XN AND YN YOU CAN WRITE THE ABOVE AS
YN=(XN+3)^2-6(XN+3)+10......OR IN CONVENTIONAL FORM WE CAN WRITE THIS AS
Y=(X+3)^2-6(X+3)+10=X^2+9+6X-6X-18+10=X^2+1...OR
Y=X^2+1.....YOU CAN CHECK ALL THE ORDERED PAIRS AND THEY SATISFY THIS EQUATION
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