# How do you transform y=1/x to y= (3x-2)/(x+1) ?

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The answer says it's a reflection in the x-axis, stretching by a scale factor of 5 and translate it by (-1.3) but how do I do it?

The answer says it's a reflection in the x-axis, stretching by a scale factor of 5 and translate it by (-1.3) but how do I do it?

##### 1 Answer

That's an interesting question! I've never done something like it before, but I've figured out how to do it.

A typical transformation of a function

#y=af[b(x-c)] + d#

where

#a# stretches/squishes the function up and down,#b# stretches/squishes the function left to right,#c# shifts the function left to right, and#d# shifts the function up and down.

In this form, a change in any of

Thus, if we can write a given "final transformation" in this form, we can extract the stretches and shifts from it directly. That's what we'll try to do.

For the function

#y=a/(b(x-c)) + d#

Can we write

Yes we can!

The first (and hardest) thing to do is to express the numerator is a way that uses

#y=(3(x+color(blue)1-color(red)1)-2)/(x+1)#

#color(white)y=(3(x+color(blue)1)-color(red)3-2)/(x+1)#

#color(white)y=(3(x+1)-5)/(x+1)#

Why did we do that? Because, when we split the function into two fractions, like this:

#y=(3(x+1))/(x+1) -(5)/(x+1)#

the first fraction has a *cancellation* we can do:

#y=(3cancel((x+1)))/cancel(x+1) -(5)/(x+1)#

#color(white)y=3 - 5/(x+1)#

And now, if we reorder the tems, we get

#y=(-5)/(x+1) + 3#

And hey, look—this is the form we were aiming for! This form tells us:

#a=–5#

#b=1#

#c=–1#

#d=3#

In other words, to translate

*(Since #b=1#, there is no left-to-right stretch or squish.)*

These instructions can be expressed the way your answer does it:

- reflect around the
#x# -axis,- stretch (vertically) with a scale factor of 5, and
- translate by
#(–1, 3)# .

Hope this helps!