How do you graph #y = 3/4 cos 3 (x - 2) - 1#?

We will construct this graph in the following sequence of steps:

1. #y=cos(x)#
   graph{cos(x) [-5, 5, -3, 3]}

2. #y=cos(3x)#
   This transformation squeezes the graph horizontally along the X-axis towards Y-axis by a factor of #3# because, if point #(a,b)# belongs to graph of #y=f(x)# (that is, #a# and #b# satisfy #b=f(a)# equation) then point #(a/K,b)# belongs to graph #y=f(Kx)# since #f(Ka/K)=f(a)=b#
   graph{cos(3(x)) [-5, 5, -3, 3]}

3. #y=3/4cos(3x)#
   This transformation stretches the graph vertically along the Y-axis by a factor of #3/4# because, if point #(a,b)# belongs to graph of #y=f(x)# (that is, #a# and #b# satisfy #b=f(a)# equation) then point #(a,Kb)# belongs to graph #y=Kf(x)# since #Kf(a)=Kb#
   graph{3/4cos(3(x)) [-5, 5, -3, 3]}

4. #y=3/4cos(3(x-2))#
   This transformation shifts the graph horizontally along the X-axis by #2# to the right because, if point #(a,b)# belongs to graph of #y=f(x)# (that is, #a# and #b# satisfy #b=f(a)# equation) then point #(a+delta,b)# belongs to graph #y=f(x-delta)# since #f(a+delta-delta)=f(a)=b#
   graph{3/4cos(3(x-2)) [-5, 5, -3, 3]}

5. #y=3/4cos(3(x-2))-1#
   This transformation shifts the graph vertically along the Y-axis by #1# down because, if point #(a,b)# belongs to graph of #y=f(x)# (that is, #a# and #b# satisfy #b=f(a)# equation) then point #(a,b-delta)# belongs to graph #y=f(x)-delta# since #f(a)-delta=b-delta#
   graph{3/4cos(3(x-2))-1 [-5, 5, -3, 3]}