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 January 25th, 2012, 08:06 AM #1 Newbie   Joined: Jan 2012 Posts: 3 Thanks: 0 turning points and nature of turning points hi, i need to know how to work this out: find the turning points and nature of turning points for: y=(3x-1)^2-4 for this i have expanded the brackets and got 9x^2-6x+1 then i minus 4 to get 9x^2-6x-3=0 however now im not sure what to do from here and this is where i require some help please. could you explain what i have to do next so i understand it and show me step by step please? thanks.
January 25th, 2012, 08:41 AM   #2
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Re: turning points and nature of turning points

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 Originally Posted by harry buckle hi, find the turning points and nature of turning points for: y=(3x-1)^2-4 for this i have expanded the brackets and got 9x^2-6x+1 then i minus 4 to get 9x^2-6x-3=0
What is a "turning point"? That is, what do YOU think a turning point is? Also, why did you expand the parentheses and subtract 4?

January 25th, 2012, 08:54 AM   #3
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Re: turning points and nature of turning points

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Originally Posted by mrtwhs
Quote:
 Originally Posted by harry buckle hi, find the turning points and nature of turning points for: y=(3x-1)^2-4 for this i have expanded the brackets and got 9x^2-6x+1 then i minus 4 to get 9x^2-6x-3=0
What is a "turning point"? That is, what do YOU think a turning point is? Also, why did you expand the parentheses and subtract 4?
its the maximum and minimum points on a curve. i did box method to work out (3x-1)^2, wouldnt i do that??

 January 25th, 2012, 09:10 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 5,738 Thanks: 342 Re: turning points and nature of turning points Harry, do you realise that 9x^2 - 6x - 3 = 0 is same as 3x^2 - 2x - 1 = 0 ? Can you factor that?
 January 25th, 2012, 09:21 AM #5 Math Team   Joined: Apr 2010 Posts: 2,767 Thanks: 353 Re: turning points and nature of turning points Only the valid values for x and the x - coordinate of the top are the same there. (3x-1)^2 - 4 has a minimum for 3x - 1 = 0 and is equal to (3x-1)^2 - 2^2 = (3x - 1 - 2)(3x - 1 + 2) = 3(x - 1)(3x + 1)
 January 25th, 2012, 09:22 AM #6 Newbie   Joined: Jan 2012 Posts: 3 Thanks: 0 Re: turning points and nature of turning points would x = 1 and x= -0.3?
 January 25th, 2012, 11:58 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 433 Math Focus: Calculus/ODEs Re: turning points and nature of turning points Are you expected to use differential calculus? I ask because normally finding relative extrema is done in the first semester of calculus, but a parabolic function in vertex form has its turning point given to you. You were given: $y=(3x-1)^2-4$ We merely need to factor out the coefficient of x, which is being squared, to write: $y=9$$x-\frac{1}{3}$$^2-4$ A parabola in the form: $y=a(x-h)^2+k$ has its turning point at (h,k) and this turning point will be a minimum if a > 0 and a maximum if a < 0.
 January 25th, 2012, 07:16 PM #8 Global Moderator   Joined: Dec 2006 Posts: 14,751 Thanks: 889 It's a bit odd to ask for the turning points (plural) when there is only one. There was no reason to add "= 0". By first considering the corresponding problem for y = (3x - 1)², it becomes "obvious" that (3x - 1)² - 4 has a minimum at (1/3, -4) and no maximum. Since y = (3x - 1)² - 2² = (3x - 1 - 2)(3x - 1 + 2), the x-axis intercepts are at x = -1/3 and x = 1. By symmetry, the x-coordinate of the turning point is the mean of these values, which is 1/3. The corresponding y-coordinate is -4.

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