With the equation y=abs(-x+3)-4y=|−x+3|−4, we know that the term inside the absolute value mark will never be less than zero, which means the smallest yy can be is y=-4y=−4. This value is achieved with abs(-x+3)=0|−x+3|=0, putting x=3x=3. So that is our first point, (3, -4)(3,−4).
We're dealing with xx terms (versus x^2x2 or sqrtx√x or any other form of xx), so the graph coming off (3,-4)(3,−4) will be lines. Since there the coefficient of the xx term is 1, the slope of those lines will be 1 and -1−1 (with the x term inside the absolute value, we look at both +-1±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)(4,−3) and (2, -3)(2,−3) - and we can plug in those values to prove it:
y=abs(-x+3)-4y=|−x+3|−4
-3=abs(-4+3)-4−3=|−4+3|−4
-3=abs(-1)-4−3=|−1|−4
-3=1-4−3=1−4
-3=-3−3=−3
and
-3=abs(-2+3)-4−3=|−2+3|−4
-3=abs(1)-4−3=|1|−4
-3=1-4−3=1−4
-3=-3−3=−3
The graph itself will look like this:
graph{abs(-x+3)-4 [-10, 10, -5, 5]}
