Unit 1, Lesson 9
What’s the Equation?
Algebra 2
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On the grid, draw a square of side length 1 in the lower left corner. Draw another square of side length 1 that shares a side directly above the first square. Next, add a 2-by-2 square, with one side along the sides of both of the first two squares. Next, add a square with one side that goes along the sides of the previous two squares you created. Then do it again.
Pause here for your teacher to check your work.
The model we use for a function can depend on what we want to do.
For example, an 8-by-10 piece of paper has area 80 square inches. Picture a set of pieces of paper, each half the length and half the width of the previous piece.
A set of 4 pieces of paper, landscape. The first is the largest paper. The second, is half the length and half the width of the first. The third and fourth, are each half the length and half the width of the previous piece of paper.
Define the sequence so that is the area, in square inches, of the piece. Each new area is the previous area, so we can define recursively like this:
But for -values larger than 5 or 6, the model isn't realistic since cutting a sheet of paper accurately when it is less than of a square inch isn't something we can do well with a pair of scissors. We can see this by looking at the graph of shown here.
If we wanted to define the term of , it's helpful to first notice that the area of the piece is given by , where there are factors of . Then we can write this definition:
Graph of a discrete function, A, origin “O”. Horizontal axis, “piece number”, with scale 0 to 6, by 1’s. Vertical axis, “area in square inches”, with scale 0 to 100, by 20’s. Coordinates plotted 1 comma 80, 2 comma 20, 3 comma 5, 4 comma1 point 2 5, 5 comma 0 point 3 1 2 5 , and 6 comma 0 point 0 7 8 1 2 5 .
We can use this definition to calculate a value of without having to calculate all the values that came before it. But since there are fewer than 10 values that make sense for since we can’t cut very tiny pieces using scissors, in this situation we could use the first definition we found to calculate different values of .