Move Function Right

[mathdad]

Given f(x) = (x - 1)^4, we move the final graph one unit to the right. This is indicated by the negative sign inside the parentheses. Sullivan does not explain why the graph is shifted right when there is a negative sign inside the parentheses. Why is this the case?

 

[Dr.Peterson]

There are several good ways to explain this, which work for different people. I'll offer one.

Consider a point on the graph of y = x^4, say (2, 16).

Now consider the function y = (x - 1)^4. To get the same y = 16, we need the number in the parentheses to be 2. But that means x - 1 = 2, so x = 3: this function subtracts 1 from x, so x has to be 1 more in order to get the same y. The point (3, 16) will be on the graph of this new function: our original point (2, 16), moved 1 unit to the right.

In general, what is done to x on the "inside" of a transformed function must be reversed to find the new x.

 

[mathdad]

There are several good ways to explain this, which work for different people. I'll offer one.

Consider a point on the graph of y = x^4, say (2, 16).

Now consider the function y = (x - 1)^4. To get the same y = 16, we need the number in the parentheses to be 2. But that means x - 1 = 2, so x = 3: this function subtracts 1 from x, so x has to be 1 more in order to get the same y. The point (3, 16) will be on the graph of this new function: our original point (2, 16), moved 1 unit to the right.

In general, what is done to x on the "inside" of a transformed function must be reversed to find the new x.

I had to read it twice to understand your reply.

 

[Harry_the_cat]

Set up a similar table to the one I suggested on your other post.