How do you find the vertex, directrix and focus of #y= -3(x+1)^2 - 4#?

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Answer 1 · 2017-07-28T11:44:20

Vertex is at # (-1 , -4) # , directrix is # y = -47/12# ,
focus is at # (-1, -49/12)#

#y = -3(x+1)^2 -4 #. Comparing with the vetex form of equation

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

# h= -1 , k = -4 , a =-3# . Vertex is at # (-1 , -4) # . Since

#a < 0# , the parabola opens downwards. Vertex is at equidistance

from focus and directrix situated downside and upper side of the

vertex. Distance of the directrix from vertex is

#d= 1/(4|a|) = 1/ (4*3)= 1/12 :. # directrix is # y = (-4 + 1/12)# or

#y = -47/12# .

Focus is at #( -1 , (-4 - 1/12)) or (-1, -49/12)#

graph{-3(x+1)^2-4 [-40, 40, -20, 20]} [Ans]