Question 884252: hey!
I am having a lot of trouble with my year 12 maths and was wondering if anyone was able to assist!
the problem solving question I have been given with, shows a table of variables being
x | 0 | 1 | 2 | 3 | 4 |
y | -3 | 0 | 5 | 12 | 21 |
for this, it then explains the first and second difference between these numbers
I understand this as the first difference is
between -3 and 5 the difference = 3 between and 0 and 5 the difference = 5, between 5 and 12 the difference = 7, and between 12 and 21 the difference = 9
the second difference then being the difference between these first difference numbers,
3 and 5 = 2
5 and 7 = 2
7 and 9 = 2
I also understand these are all constant.
so, the question has given us the rule of y=ax^2 + bx + c which I know is a general function
it then explains that the first row of entries is obtained by substituting the x values into the rule and said that when x = 1 you just substitute that into the equation. being -> y=a(1) + b(1) + c
which then simplifies into a + b + c
the next part then says that the second row of entries (or the first difference in the pattern) is obtained by subtracting consecutive entries in the first row
my first question being - what and where are the consecutive entries ?
my second question being that the example then substitutes to
4a + 2b + c - (a + b + c) = 3a + b
I am unsure of where these substitutions (being the 4a and 2b) have come from ?
sorry for the very long explanation but I would be very grateful for an answer!
thank you for you're time
:)
Found 2 solutions by Fombitz, rothauserc:
Answer by Fombitz(32388)   (Show Source):
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! we are dealing with a quadratic sequence and we work with the y values
y | -3 | 0 | 5 | 12 | 21 |
first difference is 3 5 7 9 (subtract first y value from second y value, second y value from third y value, ....)
second difference is 2 2 2 2
since the constant is 2, we know that we have a n^2 term in the expression for the nth term in the quadratic sequence
again, we look at the y values
y | -3 | 0 | 5 | 12 | 21 |
nth 1 2 3 4 5
n^2 1 4 9 16 25
subtract n^2 from y values
-4 -4 -4 -4 -4
the nth term in the geometric sequence is n^2 - 4
we can work this from another direction, we know
-3 = a(1^2) +b(1) +c
0 = a(0^2) +b(0) +c
5 = a(5^2) +b(5) +c
or
-3 = a +b +c
0 = c
5 = 25a + 5b +c
we know c = 0
-3 = a +b
1 = 5a +b
this can be solved for a and b
a = 1, b = -4
y = x^2 -4x is the quadratic form
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