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

1 Answer
Binayaka C.
Oct 5, 2016

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]

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