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May 17th, 2017, 09:49 PM  #1 
Newbie Joined: May 2017 From: Surrey Posts: 1 Thanks: 0  Plz help asap need help now (function transformations)
What happens to the points (12, 6) in the functions: Y=√f(x) + 2 (f and (x) are both square rooted) y=1/3f(3(x+2))4? 
June 2nd, 2017, 06:52 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,820 Thanks: 750  Quote:
Now, when you refer to the point (12, 6) do you mean that (12, 6) is on the graph of y= f(x) so that f(12)= 6? If so then f(x)+ 2= 4 and √(4) is not a real number. Quote:
Assuming, as above, that you mean that f(12)= 6, to evaluate f(3(x+ 2)) with only that information, must have 3(x+ 2)= 12 so that x+2= 4 and x= 2. In that case, if you intend (1/3)(f(3(x+2)) 4= that is (1/3)(6)+ 4= 2+ 4= 6. If you intend 1/(3f(3(x+2) 4 that is 1/(1+ 4= 4+ 1/18. If you intend 1/(3f(3(x+2) 4) that is 1/(18+ 4)= 1/14.  

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