Question

What are the vertex, focus and directrix of `y=-x^2+4x+1`?

Answer 1

Vertex is at `(2,5)`: Equation of directrix is `y=21/5`: Focus is at `(2,19/4)`

Explanation:

`y= -x^2+4x+1 = -(x^2-4x+4)+4+1 = -(x-2)^2+5`. Comparing with the standard equation in vertex form `a(x-h)^2+k`,where `(h,k)` is the vertex, we get vertex at `(2,5)`.

Since `a=-1`,the parabola opens down and the directrix is at backside of vertex. Vertex is at equidistance from directrix(d) and focus. We know `d=1/(4|a|) or d=1/4`. So the equation of directrix is `y=(5+1/4)=21/5`.
The focus is at `(2, (5-1/4)) or (2,19/4)` graph{-x^2+4x+1 [-20, 20, -10, 10]}[Ans]