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January 25th, 2012, 08:06 AM  #1 
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=(3x1)^24 for this i have expanded the brackets and got 9x^26x+1 then i minus 4 to get 9x^26x3=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  
Senior Member Joined: Feb 2010 Posts: 395 Thanks: 15  Re: turning points and nature of turning points Quote:
 
January 25th, 2012, 08:54 AM  #3  
Joined: Jan 2012 Posts: 3 Thanks: 0  Re: turning points and nature of turning points Quote:
 
January 25th, 2012, 09:10 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 4,049 Thanks: 217  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: 2,535 Thanks: 289  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. (3x1)^2  4 has a minimum for 3x  1 = 0 and is equal to (3x1)^2  2^2 = (3x  1  2)(3x  1 + 2) = 3(x  1)(3x + 1) 
January 25th, 2012, 09:22 AM  #6 
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 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,056 Thanks: 386 Math Focus: The calculus  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: We merely need to factor out the coefficient of x, which is being squared, to write: A parabola in the form: 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: 11,826 Thanks: 398 
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 xaxis intercepts are at x = 1/3 and x = 1. By symmetry, the xcoordinate of the turning point is the mean of these values, which is 1/3. The corresponding ycoordinate is 4. 

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