How do you find the vertex, focus, and directrix of the parabola #y=-1/6(x^2+4x-2)#?

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Answer 1 · 2017-08-20T12:19:35

Vertex is at # (-2,1) # , the directrix is # y =2.5 # and
focus is at # (-2, -0.5) #

# y = -1/6( x^2+4x-2) or y = -1/6( x^2+4x) +1/3# or

# y = -1/6( x^2+4x+4) +4/6+1/3# or

# y = -1/6( x+2)^2+1 # Comparing with standard vertex form of

equation #y = a (x-h)^2+k ; (h,k) # being vertex , we find here

#h= -2 ,k =1 ; a = -1/6 #. so vertex is at # (-2,1) # since #a <0#

the parabola opens down. Vertex is at midway between directrix

and focus , directrix is above the vertex and focus is below the

vertex. We know distance of directrix from vertex is # d= 1/(4|a|)#

or #d= 1/(4*1/6) or d = 3/2 or d =1.5# . the directrix is at

# y = 1+1.5 or y =2.5# and focus is at

# (-2, (1-1.5) or (-2, -0.5) #

graph{-1/6(x^2+4x-2) [-10, 10, -5, 5]} [Ans]