How do you graph y = |-x+3| - 4?

See below:

Explanation:

With the equation $y = \left\mid - x + 3 \right\mid - 4$, 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 $\left\mid - x + 3 \right\mid = 0$, putting $x = 3$. So that is our first point, $\left(3 , - 4\right)$.

We're dealing with $x$ terms (versus ${x}^{2}$ or $\sqrt{x}$ or any other form of $x$), so the graph coming off $\left(3 , - 4\right)$ 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 $\pm 1$).

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 $\left(4 , - 3\right)$ and $\left(2 , - 3\right)$ - and we can plug in those values to prove it:

$y = \left\mid - x + 3 \right\mid - 4$

$- 3 = \left\mid - 4 + 3 \right\mid - 4$

$- 3 = \left\mid - 1 \right\mid - 4$

$- 3 = 1 - 4$

$- 3 = - 3$

and

$- 3 = \left\mid - 2 + 3 \right\mid - 4$

$- 3 = \left\mid 1 \right\mid - 4$

$- 3 = 1 - 4$

$- 3 = - 3$

The graph itself will look like this:

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