We are given #y=(x-2)^2(x+1)(x+4)#

To find the gradient we must differentiate, but as our equation is in the form #y=u*v*w#, where #u#, #v# and #w# are functions of #x#. We must use the product rule where:

#(dy)/(dx)=(u'*v*w)+(u*v'*w)+(u*v*w')#

In this case, #y=u^2*v*w#, where:

#u=(x-2)#,

#v=(x+1)#, and

#w=(x+4)#

#(dy)/(du)=2u#

#(dy)/(dv)=1#

#(dy)/(dw)=1#

#(du)/(dx)= 1#

#(dv)/(dx)= 1#

#(dw)/(dx)= 1#

So, #(dy)/(dx)=2((x-2)(x+1)(x+4))+(x-2)^2(1)(x+4)+(x-2)^2(x+1)(1)#

Putting our values for #x=1# in gives us #2(1-2)(1+1)(1+4)+(1-2)^2(1+4)+(1-2)^2(1+1)=2(-1)(2)(5)+(-1)^2(5)+(-1)^2(2)=-13#

As #(dy)/(dx)# is negative when #x=1#, the gradient is negative and therefore decreasing.