Adding a horizontal compression to a graph that is already horizontally compressed?

[mint12]

I need to add a horizontal compression by a factor of 1/4, then a reflection in the y-axis, followed by 2 units down.

Original equation is y = 3x - 6

I understand how to do expansions, compressions, translations, etc, but I don't understand how to add a horizontal compression of 1/4 when the original equation already shows a horizontal compression of 1/3. Do I multiply them together to get a horizontal compression of 1/12?

I am also wondering what would happen if the original equation was y = 1/3x - 6. How do I add a horizontal compression to an equation that already has a horizontal expansion?

 

[fcabanski]

y=3x-6 is the function y=x stretched by a factor of 3 and shifted down 6 units.

The rules (http://www.ltcconline.net/greenl/courses/103a/functions/shift.htm) apply to functions, not just to the x term of a function.

c*f(x) is the function stretched c units if c>0, and the function compressed c units if c<0. Note it's c times the whole function, not just times the x portion.

If f(x) = 3x-6, then to stretch/compress the function c units you must multiply the entire function by c. c*(3x-6).

f(x+c) is a shift of the original function c units to the left. So y=3x-6 shifted c units left is y =3(x+c) -6 = 3x+3c-6.

f(x) + c is the function shifted c units up (or down if c <0). So y=3x-6 shifted c units up is y = 3x - 6 + c (c>0).

I need to add a horizontal compression by a factor of 1/4, then a reflection in the y-axis, followed by 2 units down.

Original equation is y = 3x - 6

I understand how to do expansions, compressions, translations, etc, but I don't understand how to add a horizontal compression of 1/4 when the original equation already shows a horizontal compression of 1/3. Do I multiply them together to get a horizontal compression of 1/12?

I am also wondering what would happen if the original equation was y = 1/3x - 6. How do I add a horizontal compression to an equation that already has a horizontal expansion?

 

[mint12]

I don't recall saying I understood anything well, what I said was that I understood that part(certainly not well or I wouldn't be posting here) and that I didn't understand this part we're talking about now right? I understand exactly as well as I think I do.

 

[fcabanski]

The original equation contains no stretching or compressing. y=3x-6 is the original equation. It is what it is.

y=3x-6 contains no compression by 1/3. If compared to y=x it contains a stretch by a factor of 3.