With the equation #y=abs(-x+3)-4#, we know that the term inside the absolute value mark will never be less than zero, which means the smallest #y# can be is #y=-4#. This value is achieved with #abs(-x+3)=0#, putting #x=3#. So that is our first point, #(3, -4)#.

We're dealing with #x# terms (versus #x^2# or #sqrtx# or any other form of #x#), so the graph coming off #(3,-4)# will be lines. Since there the coefficient of the #x# term is 1, the slope of those lines will be 1 and #-1# (with the x term inside the absolute value, we look at both #+-1#).

The standard graph for an absolute value graph is for it to be in the shape of a V. And so we can expect our graph to have points #(4, -3)# and #(2, -3)# - and we can plug in those values to prove it:

#y=abs(-x+3)-4#

#-3=abs(-4+3)-4#

#-3=abs(-1)-4#

#-3=1-4#

#-3=-3#

and

#-3=abs(-2+3)-4#

#-3=abs(1)-4#

#-3=1-4#

#-3=-3#

The graph itself will look like this:

graph{abs(-x+3)-4 [-10, 10, -5, 5]}