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The task was worded this way:

Suppose that $f(x) = \frac {1}{2}(x-1)^2 - 3$.

State exactly how the graph of $y = f(x) $ will be transformed into:

$ y = (3x-1)^2 +1 $

The answer I provided was:

  • Firstly, it will be translated up for 4 units;
  • Secondly, it will be vertically stretched by a factor of 2;
  • Thirdly, it will be horizontally compressed by a factor of 3;
  • Finally, it will be shifted to the left by $ \frac {2}{3} $ units;

Teacher said that it was not done in the right order, the correct one being:

  • Firstly, it will be shifted to the left by $ \frac {2}{3} $ units;
  • Secondly, it will be vertically stretched by a factor of 2;
  • Thirdly, it will be horizontally compressed by a factor of 3;
  • Finally, it will be translated up for 4 units;

As can be seen, in the correct answer the first and the final records are swapped.

I wonder how much the correct order really matters. I tried to trace the transformations in this task firstly according to the order I provided and then according to the correct order. It seems that the final result was still the same.

Is there any situation when the difference in order would be critical?

(Some vivid example would be appreciated).

brilliant
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1 Answers1

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We have $f(x)=\frac12(x-1)^2-3$, and let $g(x)=(3x-1)^2+1$.

Let's see what we get if we follow your sequence of transformations:

  1. Translation by $\binom{0}{4}$ so add $4$ to the whole expression and get $$\frac12(x-1)^2+1$$
  2. Vertical stretch by factor $2$, so multiply the whole expression by $2$ and get $$(x-1)^2+2$$
  3. Horizontal compression by factor $3$, so replace every $x$ term with $3x$ and get $$(3x-1)^2+2$$
  4. Shift to the left by $\frac23$ units, so replace every $x$ term by $(x+\frac23)$ and get $$\left(3(x+\frac23)-1\right)^2+2=(3x+1)^2+2\neq g(x)$$

Now let's see what we get if we follow your teacher's sequence of transformations:

  1. Shift to the left by $\frac23$ units, and get $$\frac12\left((x+\frac23)-1\right)^2-3=\frac12(x-\frac13)^2-3$$
  2. Vertical stretch by factor $2$, and get $$(x-\frac13)^2-6$$
  3. Horizontal compression by factor $3$, and get $$(3x-\frac13)^2-6$$
  4. Translation by $\binom{0}{4}$, and get $$(3x-\frac13)^2-2\neq g(x)$$

The correct sequence should be:

  1. Horizontal compression by factor $3$, and get $$\frac12(3x-1)^2-3$$
  2. Vertical stretch by factor $2$, and get $$(3x-1)^2-6$$
  3. Translate vertically by $\binom{0}{7}$ and get $$(3x-1)^2+1$$ as required.

Rule of thumb: start with the innermost transformation and work outwards.

Generally order does not matter if the transformations consist only of translations or only of enlargements. But if there are translations and enlargements in the same axis direction, then order matters.

So for example if you take the graph of $y=x^2$ and first stretch by factor $3$ horizontally, and then translate by $\binom{1}{0}$ you will get firstly $(\frac13x)^2$ and then $\left(\frac13(x-1)\right)^2$

But the same two transformation in reverse order would result in firstly $(x-1)^2$ and then $(\frac13x-1)^2$ and clearly these resulting expressions are not the same.

David Quinn
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  • (1) Thank you for your answer, but I am puzzled by point 2 in the first part of your answer: "Vertical stretch by factor 2, so multiply the whole expression by 2". Why do you need to multiply the whole expression? Our teacher teaches us to follow this formula: $f(x) = a \cdot f[b(x-h)]+k$ , in which increasing the value of $a$ results in a vertical stretch, increasing b results in a horizontal compression, increasing $h$ results in a horizontal shift, and increasing $k$ results in vertical shift. – brilliant Oct 08 '21 at 01:43
  • (2) So, in the expression $ \frac{1}{2}(x-1)^2+1$ the value of $a$ is $ \frac{1}{2}$, and the value of $k$ is $1$. According to that logic, if you multiply the whole expression by $2$, you increase both $a$ and $k$. In other words, besides performing the vertical stretch, you also perform a vertical shift, thus, moving the graph up one unit: https://www.desmos.com/calculator/25yruopdh4 – brilliant Oct 08 '21 at 01:43
  • to achieve a vertical stretch by factor 2 you need to multiply all the output values by 2. Therefore you need to multiply the whole expression by 2. Your teacher's formula could be a bit misleading since you seem to have misunderstood it. – David Quinn Oct 08 '21 at 08:59
  • As I have explained in my answer, yes the order does matter! The formula you have quoted must be interpreted as a sequence of transformations in a particular order. First $b$ then $h$ then $a$ then $k$ – David Quinn Oct 08 '21 at 09:18
  • Would the downvoter please add a comment as to why they have downvoted (as is expected on this site) – David Quinn Oct 08 '21 at 09:51
  • "to achieve a vertical stretch by factor 2 you need to multiply all the output values by 2. Therefore you need to multiply the whole expression by 2" - But in this case besides achieving the vertical stretch you also inadvertently move the whole graph up (because you have also increased $k$ besides $a$). Did you check the link that I gave you? It shows that very clearly. – brilliant Oct 08 '21 at 10:10
  • yes I saw it. The green graph is a stretch/enlargement of the red graph by factor 2, but the blue graph is not. You may think the green graph has also been translated upwards, but it is not - it is just twice as large as the red graph relative to the x axis – David Quinn Oct 08 '21 at 11:06
  • "You may think the green graph has also been translated upwards, but it is not - it is just twice as large as the red graph relative to the x axis" - If it were only a case of enlargement than its vertex would have stayed at the same distance relative to the x axis; however, its vertex now is twice farer away from it. – brilliant Oct 08 '21 at 12:33
  • you don't seem to understand what an enlargement is...... – David Quinn Oct 08 '21 at 14:18
  • I think you are confused about what is meant by enlargement. If a curve has a minimum point at $(1,1)$ and it is stretched by factor 2 in the $y$ direction, then the new coordinates of the minimum point will be at $(1,2)$, but that does not mean that the curve has been translated upwards as well. Every $y$ value on the curve is multiplied by 2, that's all. – David Quinn Oct 08 '21 at 17:51
  • Thank you for your explanations. You don't need to waist your time on me anymore. – brilliant Oct 08 '21 at 18:34
  • And sorry for lying about three professional mathematicians. – brilliant Oct 08 '21 at 19:05
  • OK no problem... – David Quinn Oct 08 '21 at 19:42