When it gives you a set of points of a function, and is asking for the original function, you are going to want to "plug in" the x-values and see if the y-values are produced.
For example, let's start with [tex]f(x) = (x + 1)^2[/tex].
[tex](0 + 1)^2 \stackrel{?}{=} 1[/tex]
[tex]1^2 \stackrel{?}{=} 1[/tex]
[tex]1 = 1 \,\,\checkmark[/tex]
[tex](1 + 1)^2 \stackrel{?}{=} 4[/tex]
[tex]2^2 \stackrel{?}{=} 4[/tex]
[tex]4 = 4 \,\,\checkmark[/tex]
[tex](2 + 1)^2 \stackrel{?}{=} 9[/tex]
[tex]3^2 \stackrel{?}{=} 9[/tex]
[tex]9 = 9 \,\,\checkmark[/tex]
You can see that all of the points checked out for the first function, meaning that that is our answer. However, to be sure, let's quickly check over the other choices.
[tex]f(x) = 4^x[/tex]
[tex]f(0) = 4^0 = 1 \,\,\checkmark[/tex]
[tex]f(1) = 4^1 = 4 \,\,\checkmark[/tex]
[tex]f(2) = 4^2 = 16 \neq 9[/tex]
- [tex]x = 2[/tex] does not check out when used in the second answer choice
[tex]f(x) = 3x + 1[/tex]
[tex]f(0) = 3(0) + 1 = 1 \,\,\checkmark[/tex]
[tex]f(1) = 3(1) + 1 = 4 \,\,\checkmark[/tex]
[tex]f(2) = 3(2) + 1 = 7 \neq 9[/tex]
- [tex]x = 2[/tex] does not check out when used in the third option choice
[tex]f(x) = 5x - 1[/tex]
[tex]f(0) = 5(0) - 1 = -1 \neq 1[/tex]
- [tex]x = 0[/tex] does not check out in the fourth equation
Thus, our answer remains as the first choice, or f(x) = (x + 1)².
