How do you find the vertex, focus and directrix of #y= 1/16(x-2)^2+3.75#?

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Answer 1 · 2017-12-01T12:14:35

Vertex is at # (2,3.75)#, directrix is #y= -0.25# and focus is at #(2,7.75)#

#y= 1/16(x-2)^2+3.75#

The vertex form of equation of parabola is

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

# h=2 and k = 3.75 and a=1/16#

So vertex is at # 2,3.75# .Parabola opens upward

since #a# is positive. Distance of vertex from directrix is

# d = 1/(4|a|)= 1/(4*1/16)=4 :.# directrix is at #y= (3.75-4)#

or #y=-0.25#. Vertex is at midway between focus and directrix.

so focus is at #2, (3.75+4) # or at (2,7.75)#

graph{1/16(x-2)^2+3.75 [-20, 20, -10, 10]}