June 27th, 2015, 10:34 PM  #1 
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0  Transformation
I used b/2(a) for x and f(b/2(a)) for y. I know I have to figure out by factoring or whatever to state for the basic function and how it's going to shift. Problem: f(x) = 2x^2  4x + 1 How I solved it: (4)/2(2) then 4/4 then x = 1 for y I plugged in 1 y = f( 2(1)^2  4 (1) + 1 ) y = f( 2  4 + 1 ) y = 1 My attempt to solving it by factoring: A) basic function y = x^2 (0,0), (1,1), (1,1) (for basic plot of parabola) B) then I factor the f(x) = 2 (x^2  2x) + 1 < this is where I'm stuck at but the book answer was f(x) = 2(x  1)^2  1 I know this is basic algebra, please help so I can do this transformation. another similar problem where I'm pulling my hair out is f(x) = 1/2(x)^2 + x  1 (same transformation problem) Last edited by skipjack; June 28th, 2015 at 12:08 AM. 
June 28th, 2015, 12:13 AM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
This isn't so much a transformation as just finding the vertex. I'll start where you left off: \begin{align*} &\quad\,\,2(x^2  2x) + 1\\ &= 2(x^2  2x + 1  1) + 1\\ &= 2(x^2  2x + 1)  1\\ &= 2(x  1)^2  1 \end{align*} See if you can carry on with the second problem now. 
June 28th, 2015, 12:39 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324 
Instead of b/2(a), write b/(2a). As a = 2 and b = 4, b/(2a) = 4/(4) = 1, and f(1) = 1 is correct. There is no need to try to factor! The rule is that axÂ² + bx + c â‰¡ a(x + b/(2a))Â²  bÂ²/(4a) + c. Putting in a = 2, b = 4 and c = 1 gives 2xÂ²  4x + 1 â‰¡ 2(x  1)Â²  2 + 1 â‰¡ 2(x  1)Â²  1. 
June 28th, 2015, 03:06 PM  #4 
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0 
The book asks by using transformation I know I'm making my life harder but I need to know how to do it so I can perform in testing condition 
June 28th, 2015, 03:06 PM  #5  
Newbie Joined: Nov 2014 From: Internet Posts: 8 Thanks: 0  Quote:
Seriously thank you so much  

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