What are the vertex, focus and directrix of # y=4(x-3)^2-1 #?

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Answer 1 · 2017-07-18T12:44:51

Vertex is at #(3,-1) # , focus is at #(3,-15/16)# and
directrix is #y= -1 1/16#.

# y= 4(x-3)^2-1#

Comparing with standard form of vertex form equation

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

#h=3, k=-1,a=4#. So vertex is at #(3,-1) #.

Vertex is at equidistance from focus and directrix and at opposite

sides . The distance of vertex from directrix is #d = 1/(4|a|) :. #

#d= 1/(4*4)=1/16#. since #a>0#, the parabola opens upwards and

directrix is below vertex. So directrix is #y= (-1-1/16)= -17/16=-1 1/16#

and focus is at #(3, (-1+1/16) )or (3,-15/16)#

graph{4(x-3)^2-1 [-10, 10, -5, 5]} [Ans]