Answer:
f(x) = x² +2
Step-by-step explanation:
x-values are evenly spaced, and y-values decrease to a minimum of 2, then increase again. First differences are -1, 1, 3, 5, and second differences are constant at 2. This means that a 2nd degree polynomial can be used for the rule.
The minimum of f(x) occurs at x=0, so there is no horizontal shift of the vertex of the quadratic function. f(1) -f(0) = 1, so the vertical scale factor for the quadratic is 1.
The quadratic with a vertex of (0, 2) and a vertical scale factor of 1 is ...
f(x) = 1·(x -0)² +2
f(x) = x² +2 . . . . . . simplified
_____
<em>Comment on differences</em>
"First differences" are the differences between successive "y" values when the "x" values are evenly spaced. Here, they are 2-3 = -1, 3-2 = 1, 6-3 = 3, 11-6 = 5. These are not constant, so the function is not a linear function.
"Second differences" are the differences between successive first differences. Here, they are 1-(-1) = 2, 3-1 = 2, 5-3 = 2. These are constant, so the function is a quadratic (2nd-degree). When n-th differences are constant, the sequence can be modeled by a polynomial of degree n.
__
<em>Comment on determining the rule</em>
Once you know the rule is 2nd-degree, there are a number of ways you can find out what it is. One way is to write it as ...
f(x) = ax^2 + bx + c
and fill in three different values for x and f(x). This will give you three linear equations in a, b, and c, which can be solved by any of the usual means for solving systems of linear equations.
Fortunately, this set of data includes the vertex of the function, making it easy to start with the vertex form:
f(x) = a(x -h)^2 +k
where (h, k) is the vertex (minimum, in this case), and "a" is the vertical scale factor. The value of "a" is easily determined as being the difference between f(h+1) and f(h). Here, h=0, so that is f(1) -f(0) = 3-2 = 1.