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