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

Vertex is at $\left(2 , 5\right)$: Equation of directrix is $y = \frac{21}{5}$: Focus is at $\left(2 , \frac{19}{4}\right)$
$y = - {x}^{2} + 4 x + 1 = - \left({x}^{2} - 4 x + 4\right) + 4 + 1 = - {\left(x - 2\right)}^{2} + 5$. Comparing with the standard equation in vertex form $a {\left(x - h\right)}^{2} + k$,where $\left(h , k\right)$ is the vertex, we get vertex at $\left(2 , 5\right)$.
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 $d = \frac{1}{4 | a |} \mathmr{and} d = \frac{1}{4}$. So the equation of directrix is $y = \left(5 + \frac{1}{4}\right) = \frac{21}{5}$.
The focus is at $\left(2 , \left(5 - \frac{1}{4}\right)\right) \mathmr{and} \left(2 , \frac{19}{4}\right)$ graph{-x^2+4x+1 [-20, 20, -10, 10]}[Ans]