All points on the line # y=x^3+4# will be transformed by the matrix
#bb(A)=((3,0),(0,-3))#
Any point on the line will have coordinates of the form.
#(k,k^3+4)#
Thus:
#((x'),(y'))=((3,0),(0,-3))((k),(k^3+4))#
#color(white)(888888)=((3k+0),(0-3(k^3+4)))=((3k),(-3k^3-12))#
.i.e.
#x'=3k#
#y'=-3k^3-12#
Eliminating #k#:
#x=3k=>k=x/3#
#y=-3(x/3)^3-12#
#color(blue)(y=-1/9x^3-12)#
We can check this by generating an #x# and a #y# value from the original equation, putting these through the transformation and testing the results in our equation of the image curve:
#x=1# and #y=5#
#((3,0),(0,-3))((1),(5))=((3),(-15))#
Plugging in #x# in:
#y=-1/9x^3-12#
#y=-1/9(3)^3-12=-15color(white)(88)# as expected