Question

How do you describe the transformation of `f(x)=-abs(x+4)+8` from a common function that occurs and sketch the graph?

Answer 1

The graph of `y=abs(x)` is flipped over the x-axis, moved left four, and moved up eight. graph{-abs(x+4)+8 [-20, 20, -10, 10]}

Explanation:

The negative sign in front of the stuff connected to the x term means the graph flips over the x-axis. (It turns upside down because this is the equivalent of changing the sign of all the y values.)
Remember that whatever is added to or subtracted from the independent (x) variable moves the graph in the opposite direction.
Like, if you add three, the graph moves left three units, and if you subtract a number, the graph moves right that many units.
Finally, the eight tacked onto the end moves the graph vertically.

What I teach my students to do is put a finger over all the stuff connected to the `x`, and whatever's left affects the `y` values. In this case, cover up `abs(x+4)`, and you know the signs of the `y` value become negative and move up eight.

For sketching, definitely use the vertex you found at `(-4,8)` by looking at the translations and draw an upside-down absolute value function.
The `y`-intercept is an easy point to identify by plugging `0` in for `x`: `f(0)=-abs(x+4)+8 => -abs(0+4)+8=-abs(4) +8=-4+8=4` so you have a point at `(0,4)`.

If your teacher requires it, or you just feel like being a little more precise, it's easy to find your `x`-intercepts (aka zeros or solutions) too. For those, you want to plug `0` in for the function value (`y`-value) instead of for the `x`-value.
(This makes sense because you're looking for the places where the function crosses the `x`-axis, and you know the function value there is `0`.)

If `y=0`: `0=-abs(x+4)+8`.
`-8=-abs(x+4)`
Since the whole right-hand side is an absolute value expression, just use `+-8` on the left:
`+-8=x+4`
`+-8-4=x`
`x= -12,4`
So now you have the shape of the graph and all the intercepts to sketch with!