Explanation:
In general, for arbitrary (x, y) pairs, the problem is called an "interpolation" problem. There are a variety of methods of creating interpolation polynomials, or using other functions (not polynomials) to fit a function to a set of points. Much has been written on this subject. We suspect this general case is not what you're interested in.
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For the usual sorts of tables we see in algebra problems, the relationships are usually polynomial of low degree (linear, quadratic, cubic), or exponential. There may be scale factors and/or translation involved relative to some parent function. Often, the values of x are evenly spaced, which makes the problem simpler.
<u>Polynomial relations</u>
If the x-values are evenly-spaced. then you can determine the nature of the relationship (of those listed in the previous paragraph) by looking at the differences of y-values.
"First differences" are the differences of y-values corresponding to adjacent sequential x-values. For x = 1, 2, 3, 4 and corresponding y = 3, 6, 11, 18 the "first differences" would be 6-3=3, 11-6=5, and 18-11=7. These first differences are not constant. If they were, they would indicate the relation is linear and could be described by a polynomial of first degree.
"Second differences" are the differences of the first differences. In our example, they are 5-3=2 and 7-5=2. These second differences are constant, indicating the relation can be described by a second-degree polynomial, a quadratic.
In general, if the the N-th differences are constant, the relation can be described by a polynomial of N-th degree.
You can always find the polynomial by using the given values to find its coefficients. In our example, we know the polynomial is a quadratic, so we can write it as ...
y = ax^2 +bx +c
and we can fill in values of x and y to get three equations in a, b, c:
3 = a(1^2) +b(1) +c
6 = a(2^2) +b(2) +c
11 = a(3^2) +b(3) +c
These can be solved by any of the usual methods to find (a, b, c) = (1, 0, 2), so the relation is ...
y = x^2 +2
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<u>Exponential relations</u>
If the first differences have a common ratio, that is an indication the relation is exponential. Again, you can write a general form equation for the relation, then fill in x- and y-values to find the specific coefficients. A form that may work for this is ...
y = a·b^x +c
"c" will represent the horizontal asymptote of the function. Then the initial value (for x=0) will be a+c. If the y-values have a common ratio, then c=0.
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<u>Finding missing table values</u>
Once you have found the relation, you use it to find missing table values (or any other values of interest). You do this by filling in the information that you know, then solve for the values you don't know.
Using the above example, if we want to find the y-value that corresponds to x=6, we can put 6 where x is:
y = x^2 +2
y = 6^2 +2 = 36 +2 = 38 . . . . (6, 38) is the (x, y) pair
If we want to find the x-value that corresponds to y=27, we can put 27 where y is:
27 = x^2 +2
25 = x^2 . . . . subtract 2
5 = x . . . . . . . take the square root*
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* In this example, x = -5 also corresponds to y = 27. In this example, our table uses positive values for x. In other cases, the domain of the relation may include negative values of x. You need to evaluate how the table is constructed to see if that suggests one solution or the other. In this example problem, we have the table ...
(x, y) = (1, 3), (2, 6), (3, 11), (4, 18), (__, 27), (6, __)
so it seems likely that the first blank (x) will be between 4 and 6, and the second blank (y) will be more than 27.