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(1, 50)

(2, 150)

(3, 300)

(4, 500)

(5, 750)

(6, 1050)

If (x, y)

what is the formula to find much higher x values?

Guest Aug 18, 2015

Best Answer 

 #6
avatar+1311 
+5

Miss Melody, I look and see if maybe can I do it .

I need practice in my writing so it will help for that too.

Also if I make a mistake I can fix it if you tell me.

Dragonlance  Aug 19, 2015
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6+0 Answers

 #3
avatar+17655 
+6

If you subtract successive y-terms (bottom up so that they are all positive):

150 - 50 = 100         300 - 150 = 150         500 - 300 = 200           750 - 500 = 250   1050 - 750 = 300

If these answers were all the same number, your formula would be a linear equation:  y = ax + b, but since they aren't, subtract the answers (again, subtract smaller from larger):

150 - 100 = 50     200 - 150 = 50     250 - 150 = 50     300 - 250 = 50

Since these answers are all the same and since it took two sets of subtractions, the formula will be quadratic:  ax2 + bx + c = y

(1,50)  -->  x = 1, y = 50  -->  a(1)2 + b(1) + c = 50   -->  a + b + c = 50

(2,150) -->  x = 2, y = 150 -->  a(2)2 + b(2) + c = 150   -->  4a + 2b + c = 150

(3,300) -->  x = 3, y = 300 -->  a(3)2 + b(3) + c = 300   -->  9a + 3b + c = 300

Now solve these for a, b, and c:

Combining the first two:  a + b + c = 50

                                       4a + 2b + c = 150

                   Subtracting:  3a + b = 100                (subtract bottom up)

Combining the last two:  4a + 2b + c = 150

                                        9a + 3b + c = 300

                   Subtracting:  5a + b = 150                 (subtract bottom up)

Combining these two answers:

                                     3a + b = 100

                                     5a + b = 150

                  Subtracting:  2a = 50   ===>   a = 25

Substituting back:  5(25) + b = 150   --->   b = 25

and finally, a + b + c = 50  --->   25 + 25 + c = 50   ===>   c = 0

So, the equation is:  ax2 + bx + c = y   ===>   y = 25x2 + 25x

geno3141  Aug 18, 2015
 #4
avatar+1311 
+6

This is really cool Geno. I not know how to do this until now.

Dragonlance  Aug 18, 2015
 #5
avatar+91051 
+1

Yes thank you Geno.  I would not have known how to do it either. 

I might have muffled through to an answer but it would not have been well organised like yours is. 

Thank you also Dragonlance for bringing Geno's answer to my attention!

I have put this in our "Reference material" sticky topic   threads   

 

That Reference thread really needs to be reorganised - any volunteers ? :))

Melody  Aug 19, 2015
 #6
avatar+1311 
+5
Best Answer

Miss Melody, I look and see if maybe can I do it .

I need practice in my writing so it will help for that too.

Also if I make a mistake I can fix it if you tell me.

Dragonlance  Aug 19, 2015
 #7
avatar+91051 
0

Ok Dragonlance, I will let you know if you make a mistake. :))

Melody  Aug 19, 2015
 #8
avatar+91051 
0

Dragonlance, I only just realised that you were offering to help with the Reference material thread.

That is very nice of you. Thank you.    

It is a big job, you have not been posting very long, maybe it is too big a job for you ?  

If you can do it, that would be great :)

Melody  Aug 22, 2015

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