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March 18th, 2017, 06:32 PM   #1
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Show that the function can be written as...

a) Given the function f(x)=(2x^2-12x+15)/〖(x-3)〗^2 ;x≠3, show that the function can be written as f(x)=(-3)/〖(x-3)〗^2 +2;x≠3
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March 18th, 2017, 06:46 PM   #2
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$f(x) = \dfrac{2x^2 - 12x + 15}{(x-3)^2},~x \neq 3$

$2(x-3)^2 = 2(x^2 - 6x + 9) = 2x^2 -12x + 18 $

$2x^2 - 12x + 15 = 2(x-3)^2 - 3$

$f(x) = \dfrac{ 2(x-3)^2 - 3}{(x-3)^2} = $

$\dfrac{2(x-3)^2}{(x-3)^2}-\dfrac {3}{(x-3)^2} = $

$2 -\dfrac{3}{(x-3)^2}$
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March 18th, 2017, 06:50 PM   #3
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This is a little tricky.

$\dfrac{2x^2 - 12x + 15}{(x - 3)^2} = \dfrac{2x^2 - 12x + 15 + 3 - 3}{(x - 3)^2} = \dfrac{2(x^2 - 6x + 9) - 3}{(x - 3)^2}.$

Now what?
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March 18th, 2017, 07:00 PM   #4
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$$\frac{2x^2-12x+15}{(x-3)^2}=\frac{2x^2-11x+15-x}{(x-3)^2}=\frac{2x-5}{x-3}-\frac{x}{(x-3)^2}$$

$$\frac{2x-6+1}{x-3}=2+\frac{1}{x-3},\,2+\frac{1}{x-3}-\frac{x}{(x-3)^2}=2+\frac{-3}{(x-3)^2}$$
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March 18th, 2017, 07:36 PM   #5
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@romsek thanks for this I managed to see how to do it legit like 10 minutes after posting your way was way the way I did it but missed out a few steps so thank you for showing me them
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