To graph the function `f(x) = -(x-2)^2 + 4`, we can start by recognizing that it is a quadratic function in standard form `f(x) = -a(x-h)^2 + k` where the vertex is at `(h,k)`. In this case, `a = 1`, `h = 2`, and `k = 4`, so the vertex is at `(2,4)` and the parabola opens downwards.
To graph the function, we can find a few additional key points. First, the y-intercept occurs when `x=0`, so we can evaluate `f(0) = -(0-2)^2 + 4 = -4`. Thus, the y-intercept is at `(0,-4)`.
Next, we can find the x-intercepts by solving for when `f(x) = 0`:
```
-(x-2)^2 + 4 = 0
-(x-2)^2 = -4
(x-2)^2 = 4
x-2 = ±2
x = 2 ± 2
```
So the x-intercepts are at `(0,0)` and `(4,0)`.
With these key points, we can sketch the graph of the function as follows:
```
6| *
| * *
| * *
| * *
| * (4,0) *
|* *
0|--------------------------
-2 -1 0 1 2 3 4 5
(2,4)
```
The vertex is at `(2,4)` and the parabola opens downwards. The y-intercept is at `(0,-4)` and the x-intercepts are at `(0,0)` and `(4,0)`.