Quod Erat Demonstrandum

2017/05/28

Transformation of graphs

Filed under: NSS,Teaching — johnmayhk @ 8:48 下午
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Share some points about the topic ‘transformation of graphs’. Nothing new.

Transformation of graphs in the HKDSE syllabus refers to

– translation
– reflection (with respect to the x-axis or the y-axis)
– enlargement or contraction (along the x-axis or the y-axis)

and all happen in the xy-plane.

Point 1.

Will the results of the following transformations be the same?

(a) A graph g is obtained by translating the graph of f one unit upwards and followed by reflecting the graph with respect to the x-axis.

(b) A graph g is obtained by reflecting the graph of f with respect to the x-axis and followed by translating the graph one unit upwards.

No.

It is crystal clear to see the fact when we use algebraic expressions to represent g in terms of f.

For (a), g(x)=-(f(x)+1).

For (b), g(x)=-f(x)+1.

They are not the same for sure.

From the example above, we see that sometimes the order of transformations is matter.

In fact, the results of the following transformations are the same:

(a) A graph g is obtained by translating the graph of f one unit upwards and followed by reflecting the graph with respect to the x-axis.

(c) A graph g is obtained by reflecting the graph of f with respect to the x-axis and followed by translating the graph one unit downwards.

So, markers not only look at the key words like ‘reflection, x-axis’ and ‘1 unit upwards’ in the marking scheme, but also pay attention to the order of transformations, ‘1 unit downwards’ is acceptable when following the order in (c).

Point 2.

When translating the graph of y=f(x) 1 unit to the right followed by enlarging the graph to 2 times the original along the x-axis, what will be the expression of the resulting graph?

(i) y=f(\frac{x-1}{2})

(ii) y=f(\frac{x}{2}-1)

Which is your choice? (i)?

Try to explore at

https://www.geogebra.org/m/VDrEAXu3

The answer is (ii).

That day in class, students puzzled, and I immediately used an ‘intermediate expression h(x)‘ to help them understand why,

f(x) === translate 1 unit rightwards ===> h(x)
h(x) === enlarge to 2 times the original along the x-axis ===> g(x)

hence,

h(x)=f(x-1)
g(x)=h(\frac{x}{2})

thus,

replace x by \frac{x}{2} in the expression h(x)=f(x-1), i.e.

g(x)=h(\frac{x}{2})=f(\frac{x}{2}-1)

I had added the graph of y=f(\frac{2x-3}{5}) in the Geogebra file above, students, can you describe the transformations from the graph of y=f(x) to that of y=f(\frac{2x-3}{5})?

(Hint: Let’s consider an intermediate expression h(x)=f(\frac{2}{5}x).)

Point 3.

The concept of enlargement or contraction along the x-axis or the y-axis is different from the enlargement or contraction of similar figures that students had come across in junior form; may be due to this reason, it is not easy to accept the following:

When enlarging the graph of y=x^3 to 2 times the original along the x-axis followed by enlarging the graph to 8 times the original along the y-axis, the resulting graph will completely overlap the original graph y=x^3.

Algebra tells us the following, the equation of the resulting graph is

y=8(\frac{x}{2})^3=x^3

To persuade students, I just used desoms to demonstrate that day:

https://www.desmos.com/calculator/q7bklunnjc

Point 4.

Casually, I’d set the following question in a test:

Describe the transformation from the graph of y=f(x)=(x-1)(x-2)(x+1) to that of y=g(x)=x(x-2)(x+1).

(Students, you try to solve it.)

It may be a bit tricky, but not difficult, in algebra. Notice that

f(-x)=(-x-1)(-x-2)(-x+1)=-(x+1)(x+2)(x-1)

\Rightarrow -f(-x)=(x+1)(x+2)(x-1)

\Rightarrow -f(-(x-1))=x(x+1)(x-2)

Therefore,

g(x)=-f(-(x-1))

Okay, based on the expression “-f(-(x-1))“, we are going to describe the transformations involved:

The graph of y=g(x) is obtained by translating the graph of y=f(x) 1 unit to the right, followed by reflecting it with respect to the y-axis and then reflecting it with respect to the x-axis.

Is it correct?

In fact, the statement above in red is wrong.

We know that the graph of y=f(-x) is obtained by reflecting the graph of y=f(x) with respect to the y-axis; however, the graph of y=f(-(x-1)) is not obtained by reflecting the graph of y=f(x-1) with respect to the y-axis. Instead, it is obtained by reflecting the graph of y=f(x-1) with respect to the line x=1.


(The graph of y=f(x-1) is in green, while that of y=f(-(x-1)) in orange.)

Proof?

For any point (a,b) lying on the graph of y=f(x-1), we have

b=f(a-1)=f(-(1-a))=f(-((2-a)-1)),

that is, (2-a,b) lies on the graph of y=f(-(x-1)).

Knowing that, (2-a,b) is obtained by reflecting (a,b) with respect to the line x=1. (Why?)

Thus, the whole graph of y=f(-(x-1)) can be obtained by reflecting the graph of y=f(x-1) with respect to the line x=1.

Out of syllabus?

Well, in the HKDSE syllabus, there is nothing to do with the reflection with respect to lines other than x-axis or y-axis. So, instead of reflecting with respect to the line x=1, we can use other transformations to achieve the same effect:

(a,b) ===> (-a,b) ===> (-a+2,b)

i.e. Reflecting with respect to the y-axis and then translating 2 units to the right.

All in all, the graph of y=x(x-2)(x+1) is obtained by translating the graph of y=(x-1)(x-2)(x+1) 1 unit to the right, followed by reflecting it with respect to the y-axis, then translating it 2 units to the right and finally, reflecting it with respect to the x-axis.

Let’s visualize the above at https://www.geogebra.org/m/dH2X7Dp9.

Point 5.

I really want to let students understand why, as for example, the graph of y=f(x-1) is obtained by translating that of y=f(x) 1 unit to the right as told in a meeting with CDI people, however, how poor I am to face the reality to skip most of the proofs and create some ‘formulae’ for memorization only

https://www.facebook.com/photo.php?fbid=10154631042258231&set=a.10154147274063231.1073741860.573778230&type=3&theater

Sadly.

Finally, I remember suddenly that at the very beginning of teaching of this topic, a student asked me that instead of considering constants k, what about the graph of y=k(x)f(x) or y=f(k(x))? Left as exercise to readers.

廣告

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