Pre-Calculus Pre-Calculus Math Forum

 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. Thanks from p41s3r 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 ax² + bx + c ≡ a(x + b/(2a))² - b²/(4a) + c. Putting in a = 2, b = -4 and c = 1 gives 2x² - 4x + 1 ≡ 2(x - 1)² - 2 + 1 ≡ 2(x - 1)² - 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
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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 Tags transformation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post xamdarb Algebra 1 March 13th, 2014 10:16 AM jiasyuen Algebra 5 March 7th, 2014 05:41 AM 667 Linear Algebra 0 September 21st, 2013 06:02 PM hereiam Real Analysis 6 July 26th, 2011 10:59 AM tuzzi-i Linear Algebra 2 June 10th, 2007 07:41 AM

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