Answer:
a. Explicit: f(x) = 2x² +3. Recursive: f(-3) = 21; f(x) = f(x -1) +4x -2
b. Explicit: g(x) = 3x² -2x. Recursive: g(5) = 65; g(x) = g(x -1) +6x -5
Explanation:
Given two tables of values, you want to find the explicit and recursive formulas for those values.
General approach
When the x-values are consecutive integers, as they are in these tables, it is usually useful to look at the first difference of the y-values. If those are constant, the table represents a linear function.
When the first differences are not constant, the exercise can be repeated. If the second differences are constant (as here), then the explicit formula will be a second-degree polynomial. The coefficients of the polynomial can be found different ways. We will describe one way here.
2nd degree explicit formula
The constant second differences are double the leading coefficient of the quadratic that describes the relation between x and y. Knowing this, we can see how well the 2nd degree term by itself represents the relation. To that end, we have added to the table columns for the difference between the y-value and the 2nd degree term (y1 -ax^2), and the difference of those differences (diff).
Explicit formula, Table A
The attachment shows the first second difference of the values of Table A is 4. That means our first approximation of the sequence will be (4/2)x². The next column shows us this is always 3 less than the value of y, so our explicit formula can be ...
f(x) = 2x² +3
Explicit formula, Table B
The attachment shows the first second difference of the values of Table B is 6. That means our first approximation of the sequence will be (6/2)x². Subtracting this from the value of y, we find an arithmetic sequence that has a common difference of -2, and a value of -10 when x=5.
The explicit formula for an arithmetic sequence with a first term a1 and a common difference d is ...
an = a1 +d(n -1) . . . . . . where 1 is the x-value of the first term
Our sequence of differences doesn't start with x=1, but with x=5. This means our explicit formula will be ...
g(x) = 3x² +(-10 +(-2)(x -5))
g(x) = 3x² -2x
2nd degree recursive formula
A recursive formula is one that expresses a given term as a function of previous terms of the sequence. When the sequence is 2nd degree, as here, we know the difference from one term to the next is an arithmetic sequence. Thus we can write the formula for a term as the sum of the previous term and a value that depends on the term number.
The explicit formula for the arithmetic sequence of differences has the form shown above. For the purpose here, the difference from the previous term is shown on the line above the x-value.
Recursive formula, Table A
The applicable sequence of first differences has first term -10 and common difference 4. The first of the differences of consequence corresponds to x=-2, so can be written ...
d(x) = -10 +4(x -(-2)) = 4x -2
Then the recursion relation is
f(x) = f(x -1) +d(x) = f(x -1) +4x -2
And the recursive formula is ...
f(-3) = 21; f(x) = f(x -1) +4x -2
Recursive formula, Table B
The applicable sequence of first differences has first term 31 and common difference 6. The first of the differences of consequence corresponds to x=6, so can be written ...
d(x) = 31 +6(x -6) = 6x -5
Then the recursive formula is ...
g(5) = 65; g(x) = g(x -1) +6x -5