# Graphing functions with mapping vector vs transformation vector

Teacher 1 method to graph $$y=2\sqrt[3]{2(x-1)}+4$$ says:

1. Create the transformation vector of $$<1,4>$$ and plot this point as an ordered pair.
2. Create a table of values for the parent graph of $$y=\sqrt[3]x$$ using easy, reasonable x values $$(-8, -1, 0, 1, 8)$$.
3. To this parent graph table, create 2 new columns and modify the x and y values with the stretch and compression values.
4. From the plotted ordered pair of $$(1,4)$$ move over and up according to the new modified values. This works and creates correct ordered pairs.

Teacher 2 method: 1. Create a mapping vector of $$<1/2 x+1, 2y+4>$$

1. Simply apply these to the parent graph ordered pairs.

teacher 2 method seems simpler. But then I ran into a function that doesn't seem to work with teacher 1 as follows:

$$y=2^\left(x+1\right)-3$$.

When I try to create the transformation vector of $$<-1,-3>$$ I see that this ordered pair is NOT on the graph. So applying the parent graph values to a point NOT on the graph clearly produce erroneous results.
When I use the method of creating $$ this works properly.

Why doesn't Teacher method 1 work for this simple exponential?

I think you know why method 1 didn't work for the exponential. It's because, exactly as you say, $$(-1,-3)$$ is not on the graph. Use the transformation vector $$(-1,-3)$$, but apply the parent graph values to the point $$(-1,-2)$$, which is on the graph, and see if that doesn't work.
Basically, when you strip away everything from the first equation, you're left with $$y=\root3\of x$$, which goes through the origin; but when you do the same for the second equation, you get $$y=2^x$$, which doesn't. That's the difference.