Quadratic Transformation

[Jakotheshadows]

I have a generic question concerning transformations of the basic function f(x)=x^2

Scenario: y = 2(3x - 1)^2 - 4 or 2f(3x-1)-4
I understand that the 3x implies a horizontal shrink by factor (1/3), and that the -1 is a shift to the left by one unit. I also understand that the 2 factor outside the parentheses indicates a vertical stretch by factor 2, and that the -4 is a shift down by four units. What I am unsure of is in which order all of this is supposed to be done. Particularly, I am not sure if the vertical shrink should be done first, or if the right shift should be done first. I've looked for examples of this in my textbook and they don't seem to illustrate examples of transformations requiring four steps (as in horizontal shift, vertical shift, horizontal stretch/shrink, and vertical stretch/shrink). It gives fine examples of everything except transformations involving both a horizontal stretch/shrink and a horizontal shift. Could someone please clarify which happens first?

 

[skeeter]

y = 2(3x - 1)[sup:1cs2soie]2[/sup:1cs2soie] - 4

y = 2[3(x - 1/3)][sup:1cs2soie]2[/sup:1cs2soie] - 4

y = 18(x - 1/3)[sup:1cs2soie]2[/sup:1cs2soie] - 4

now the transformation of the parent function y = x[sup:1cs2soie]2[/sup:1cs2soie] is clear ...

horizontal shift to the right 1/3 of a unit
vertical stretch by a factor of 18
vertical shift downward 4 units

 

[Jakotheshadows]

In other words, horizontal shift occurs before horizontal stretching or shrinking? I'm not entirely convinced that the process you're using is correct.

Suppose the 2 is not in the transformation. y = (3x - 1)^2 - 4,
y = 3[x - (1/3)]^2 - 4 ---> so you've just changed from a shift to the right by one unit and a horizontal shrink by factor (1/3) to a vertical stretch of factor 3 and a right shift by (1/3)? It seems like you're making it an entirely different translation by factoring it that way. This is what I'm seeing (assuming that horizontal shift occurs first as you demonstrated earlier).

y = x^2 --------------------------------------------> y = (3x - 1)^2 - 4
(-3 , 9) -------> (-2 , 9)------->(-2/3 , 9) ------->(-2/3 , 5)
(-2 , 4) -------> (-1 , 4)------->(-1/3 , 4) ------->(-1/3 , 0)
(-1 , 1) -------> (0 , 1) ------->(0 , 1) ------->(0 , -3)
(0 , 0) -------> (1 , 0) ------->(1/3 , 0) ------->(1/3 , -4) Vertex.
(1 , 1) -------> (2 , 1) ------->(2/3 , 1) ------->(2/3 , -3
(2 , 4) -------> (3 , 4) ------->(1 , 4) ------->(1 , 0)
(3 , 9) -------> (4 , 9) ------->(4/3 , 9) ------->(4/3, 5)

y = x^2 --------------------------------------------> y = 3(x - 1/3)^2 - 4
(however, when factored out this does give you the vertex: a(x - h)^2 + k, but it doesn't translate the same.)
(-3 , 9)---->(-8/3 , 9)---->(-8/3 , 27)--->(-8/3, 23)
(-2 , 4)---->(-5/3 , 4)---->(-5/3 , 12)--->(-5/3 , 8)
(-1 , 1)---->(-2/3 , 1)---->(-2/3 , 3) --->(-2/3 , -1)
(0 , 0) ---->(1/3 , 0) ---->(1/3 , 0) --->(1/3 , -4) Vertex.
(1 , 1) ---->(4/3 , 1) ---->(4/3 , 3) --->(4/3 , -1)
(2 , 4) ---->(7/3 , 4) ---->(7/3 , 12) --->(7/3 , 8)
(3 , 9) ---->(10/3, 9)---->(10/3, 9) --->(10/3 , 5)

Strangely enough, I just checked my graphing calculator and your factoring of the function I originally gave works fine. However, without a vertical stretch already in the equation, apparently a horizontal shrink can't become a vertical stretch? :?

 

[Jakotheshadows]

NEVERMIND.

I see the error of my ways.. Yes, horizontal shift always occurs first, and I forgot to square the factor that I took out of the parentheses.

(3x - 1)^2 - 4 = 9(x - 1/3)^2 - 4

That was the significance that I missed when you put the brackets [ ] around the two factors when you took 3 out the first time, with the exponent outside the brackets I realized that I couldn't remove a factor from a squared parentheses without squaring it first. I <3 math.

 

[Denis]

NEVERMIND.
I see the error of my ways..

Go stand in the corner :shock: