How do you find the vertex, directrix and focus of #y=-(1/4)x^2+2x-5#?

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Answer 1 · 2017-07-25T12:45:10

Vertex is at #(4,-1)#, directrix is # y =0# and focus is at #(4,-2)#

# y = -(1/4) x^2 +2x - 5 #. Comparing with #ax^2+bx+c# we

get # a= -1/4 , b =2 ,c =-5 #, x- coordinate of vertex is

#x= -b/(2a) = (-2)/-(2*1/4) =4 # , y- coordinate of vertex is

obtained by putting #x=4# in the equation,

# y = -1/4*4^2+2*4-5 = -1 :.# Vertex is at #(4,-1)#.

Vertex is at equidistance from directrix and focus situated at

opposite side of parabola. we know distance of directrix from

vertex is #d = 1/(4|a| )= 1/(4* 1/4) =1# . Since #a < 0# the parabola

opens downward. So directrix is #y=(-1+1) or y =0#

Focus is at # (4, (-1-1)) or (4,-2)#

graph{-1/4(x^2)+2x-5 [-10, 10, -5, 5]} [Ans]