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]}
