Help with identifying transformations (1 Viewer)

[moderntortoisecat]

Given y= - 1/4-2x +2. How can i find the transformations from the original function 1/x

 

[askit]

Im assuming you mean -1/(4-2x) + 2, which has 4-2x on the denominator and + 2 by itself. If thats the case, the original 1/x can be reflected upon the y axis hence: 1/-x, then horizontally dilated by a factor of 1/2 hence: 1/-2x, then translated right 2 hence 1/(4-2x) --> it helps to write the original function as -1/2(2-x) + 2 to see the real translation --> then reflected in the x axis, so -1/(4-2x) and then shifted vertically up by 2 so -1/(4-2x) + 2. Its been a while since I've done transformations, but the general gist is to do dilations then translations or vis versa. I wouldn't interchangeably use then because some translations + dilations don't commute (forgot which ones). There may be a quicker way to do the reflections, since reflecting in x/y for 1/x is basically the same, but yea hope this helps

 

[moderntortoisecat]

Thank you!

 

[ISAM77]

with a bit of algerba you can rewrite the equation as: y-2 = 1 / [2(x-2)]

y = 1/x is the starting graph
y-2 = 1/x moves the graph up 2
y-2 = 1 / 2x is a horizontal dilation by 1/2.
y-2 = 1 / [2(x-2)] moves the graph right 2.

You can put the 4 equations above into Geo Gebra, to see exactly how it moves. Should be helpful.

 

[Luukas.2]

@ISAM77 's form is the general form:

​

makes it easy to sketch...

• is the horizontal asymptote (or, in the general case, )
• is the vertical asymptote (or, in the general case, )
• the numerator of the fraction is positive (), and so the hyperbola looks like relative to its asymptotes (if , it looks like )

The shifts being 2 up and 2 to the right are now clear as that's what's happened to the asymptotes.