My Math Forum Plz help asap need help now (function transformations)

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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
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Quote:
 Originally Posted by subag24 What happens to the points (12, -6) in the functions: Y=√f(x) + 2 (f and (x) are both square rooted)
Then you should have written y= √(f(x)+ 2).
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:
 y=-1/3f(3(x+2))-4?
Another ambiguity. Do you mean 1/(3f) or (1/3)f? Do you mean 1/(3f- 4) or 1/(3f)- 4?

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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