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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.

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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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

here is the answer on q2e

mm2-assessrep12.pdf

mm2-assessrep12.pdf

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

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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