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October 27th, 2016, 11:08 PM   #1
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Functions

Sketch f(x) = [2/(3x - 1)] + 3

I know that the parent graph of the function is f(x) = 1/x and when I express the function of transformed graph, it is

f(x) = [2/3(x - 1/3)] + 3

I interpreted as the graph of f(x) = 1/x is moved 1/3 units to the right and it is narrower by 3 units, then it is moved 3 units upwards. Now my problem is, what effect does the 2 have on the graph?

Last edited by skipjack; October 28th, 2016 at 02:09 AM.
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October 28th, 2016, 01:47 AM   #2
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Quote:
Originally Posted by jeho View Post
Sketch f(x) = [2/(3x - 1)] + 3

I know that the parent graph of the function is f(x) = 1/x and when I express the function of transformed graph, it is

f(x) = [2/3(x - 1/3)] + 3

I interpreted as the graph of f(x) = 1/x is moved 1/3 units to the right and it is narrower by 3 units, then it is moved 3 units upwards. Now my problem is, what effect does the 2 have on the graph?
Don't type "[2/3(x-1/3)]+3."

It falls into unclear/ambiguous/can't be determined, in part, because it looks as if it could be (2/3)
multiplied by (x - 1/3), (before adding the 3), which is not what you intend.

Write them as:

Sketch f(x) = 2/(3x - 1) + 3

f(x) = 2/[3(x - 1/3)] + 3


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Take the 3 in the denominator out with the 2 as a fraction:

f(x) = (2/3)*[1/(x - 1/3)] + 3


$\displaystyle f(x) \ = \ \dfrac{2}{3}\left(\dfrac{1}{x \ - \ \tfrac13}\right) + \ 3$


Then, you would not state it is "narrower by 3 units". Because 2/3 < 1, you state that the graph has a compression factor of 2/3.

And you're correct that it has a horizontal right shift of 1/3 unit and a vertical up shift of 3 units.

Last edited by skipjack; October 28th, 2016 at 02:09 AM.
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October 28th, 2016, 02:34 AM   #3
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It makes sense now.
Thnaks very much Sir
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October 28th, 2016, 03:08 AM   #4
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That should be a vertical compression factor of 2/3, applied prior to the vertical shift.
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