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 June 27th, 2015, 09:34 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 27th, 2015 at 11:08 PM.
 June 27th, 2015, 11:13 PM #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. Thanks from p41s3r
 June 27th, 2015, 11:39 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,941 Thanks: 2210 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, 02:06 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, 02:06 PM   #5
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Quote:
 Originally Posted by Azzajazz 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.
Thank you so much, I still don't get it but I'm going to study it till my head explodes.

Seriously thank you so much

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