Question

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

Answer 1

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

Explanation:

`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]}