(a) If the child function is of the form Af(B(x-C))+D then the new domain can be calculated by dividing the ends of the domain divided by B plus C.... This is slightly beyond the basics of function transformation so if the form Af(B(x-C))+D is not yet understood or terms like shift and dilate are not fully understood then I recommend a session with a WyzAnt math tutor.
In the case of f(2(x-3)) then B = 2 and C = 3 (not -3). So 1/2 + 3 = 3.5 and 6/2+3 = 6 therefore the new domain is [3.5,6]
(b) The new range can be found by multiplying the ends of the range by A and then adding D.
In the case of f(2(x-3)) then A = 1 and D = 0. So -3•1+0 = -3 and 5•1+0 = 5. The range stays at [-3,5].
(c) In the case of 2f(x)-3 B=1 and C=0 therefore 1/1+0 & 6/1+0 therefore the new domain is still [1,6].
(d) ... A = 2 and D=-3 therefore -3•2 - 3 = -9 and 5•2 - 3= 7.
(e) To get the proper width one must dilate by a factor of 1/5 therefore B = 5. This would place the right side of the domain at 6/5 so one must shift to the right by 9-6/5 = 7 4/5 = 7.8 therefore C = 7.8 so f(5(x-7.8)).
(f) To get the proper height you must dilate by a factor of 1/8 which will place the top of the function at 5/8 so the function must be shifted by 1 - 5/8 or 3/8 up so ... A=1/8 & D = 3/8 therefore f(x)/8+3/8.
