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:


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 $<x-1, y-3$ 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.

  • $\begingroup$ Thank you. I think I will use the second method which always works. $\endgroup$
    – user163862
    Dec 6 '18 at 17:59

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