Q&A: What’s the Intuition Behind the Order of Function Transformations?

by Justin Skycak (x.com/justinskycak) on

Cross-posted from here.

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Question

Is there a good intuitive explanation for how to think about the order of steps when graphing functions through a sequence of transformations?

Answer

One way to build intuition around this is to think about the effects on the intercepts.


Vertical Shifts/Stretches: Effect on $y$-Intercept

Suppose you want to graph $y=2\sqrt{x}-6$ using transformations.

You know the result needs to have a $y$-intercept of $-6$ (which comes from evaluating $2\sqrt{0}-6$).

  • If you stretch vertically by a factor of $2$ and then shift $6$ down, you get a $y$-intercept of $-6$ as desired. $\color{green}\checkmark$
  • If you shift $6$ down and then stretch vertically by a factor of $2,$ you get a $y$-intercept of $-12$ which is incorrect. $\color{red}\times$

So vertical stretches come before vertical shifts.


Horizontal Shifts/Shrinks: Effect on $x$-Intercept

Suppose you want to graph $y=\sqrt{2x-6}$ using transformations.

You know the result needs to have an $x$-intercept of $3$ (which comes from solving $0=\sqrt{2x-6}$).

  • If you shrink horizontally by a factor of $2$ and then shift $6$ right, you get an $x$-intercept of $6$ which is incorrect. $\color{red}\times$
  • If you shift $6$ right and then shrink horizontally by a factor of $2,$ you get an $x$-intercept of $3$ as desired. $\color{green}\checkmark$

So horizontal shifts come before horizontal shrinks.


More Hand-Wavy (But Potentially More Memorable) Intuition

When it comes to horizontal transformations, everything works the opposite way as vertical transformations.

The transformations themselves are opposite:

  • shifts right (the negative horizontal direction) instead of shifts up (the positive vertical direction)
  • horizontal shrinks instead of vertical stretches

So it’s kind of intuitive that the order of transformations is also opposite.


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