# transform f(x)

Hi,

I dont get this question

f (x) = 1/(2x - 4)+ 3.

A transformation T: R2 ¿ R2 that maps the graph of f to the graph of the function
g: R\{0} ¿ R, g (x) = 1/x
has rule T

(matrix)
[x ] =[a 0 ] [ x]   +[c]
[y]    [0 1]   [  y]   [d]

, where a, c and d are non-zero real numbers.
Find the values of a, c and d.
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Commented:
y = 1/(2x - 4) + 3

The matrix gives  x = ax' + c  and y = y' + d

Now choose a, c, d to transform into y' = f(x') = 1/x'

Substituting gives

y' + d  = 1/(2(ax' + c ) - 4) + 3

y'   = 1/(2ax' +2c- 4)  + 3 - d   = 1/x'

Clearly  for y' = 1/x'

2a = 1

2c- 4 = 0

3 - d = 0

Giving a = 1/2 , c = 2,   d =3
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Author Commented:
I didnt get how you got  'aa

y'   = 1/(2ax' +2c- 4)  + 3 - d   = 1/x'

Clearly  for y' = 1/x'

2a = 1
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Commented:
I take

y = 1/(2x - 4) + 3

Then replace x and y with  x = ax' + c  and y = y' + d. This comes from the matrix transform

Giving

y' + d  = 1/(2(ax' + c ) - 4) + 3

If y' = 1/x'.  I now need d=3, a=1/2, c=2
0
Author Commented:

y'=y+d
x'=ax+c

d=-3
a=2
c=-4
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Author Commented:
here is the answer on q2e
mm2-assessrep12.pdf
0
Commented:
It was unclear from your transformation matrix what was the origin variable and the new transformed one. I used for me the more intuitive interpretation, which was incorrect. Having clarified that then

y = 1/(2x - 4) + 3

The matrix gives  x' = ax + c  and y' = y + d

ie     x =(x' - c)/a    and     y= y' - d

Now choose a, c, d to transform into y' = f(x') = 1/x'

Substituting gives

y' - d  = 1/(2(x' - c)/a - 4) + 3

y'   = 1/(2x'/a - 2c/a- 4)  + 3 + d   = 1/x'

Clearly  for y' = 1/x'

2/a = 1

-2c/a- 4 = 0

3 + d = 0

Giving a = 2 , c = -4,   d = -3
0

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