Question 1012164
{{{(x+5)^2=4y }}}

{{{y=(1/4)(x+5)^2 }}}....compare to {{{y=(x-h)^2+k}}} and you see that {{{h=-5}}} and {{{k=0}}}

so, vertex is at ({{{-5}}},{{{0}}}

or, we can use  standard form and calculate coordinates of the vertex like this:

{{{y=(1/4)(x^2+10x+25) }}}

{{{y=(1/4)x^2+(1/4)10x+(1/4)25 }}}

{{{y=(1/4)x^2+(5/2)x+(25/4) }}}

the x-coordinate of the focus is:

{{{x = -b /(2a)}}} 

{{{x = -(5/2)/ (2(1/4))}}} 

{{{x = -(5/2)/ (1/2)}}} 

{{{x = -(10/2)}}} 

{{{x = -5}}} 

then, y-coordinate is

{{{y=(1/4)(-5)^2+(5/2)(-5)+(25/4) }}}

{{{y=(1/4)25-(25/2)+(25/4) }}}

{{{y=25/4-50/4+(25/4) }}}

{{{y=50/4-50/4 }}}

{{{y=0 }}}

so, we got same result for vertex


the focus is:

The focus lies on the axis of symmetry of the parabola, and has the y-coordinate {{{k+1/(4a)}}}. Because we just found the vertex to be ({{{-5}}},{{{0}}}), we know the axis of symmetry to be {{{x=-5}}}, and the focus lies on that line.

y-coordinate is at {{{1/(4a) =1/(4(1/4))=1/(4/4)=1}}}

and the coordinates of the focus are:({{{-5}}},{{{1}}})


Once you know the y=coordinate of the vertex, {{{k=0}}}, it is given by {{{y = k – p}}}, where {{{p = 1/(4a)}}}. Thus, as we calculated for the focus, above:

{{{p = 1/(4a) = 1/(4*(1/4)) = 1}}}

and the directrix is {{{y = 0 – 1 =-1}}}



{{{drawing(600, 600, -10, 10, -10, 10,
circle(-5,0,.12),circle(-5,1,.12),
locate(-5,-0.5,V(-5,0)),locate(-5,1,F(-5,0)),
 graph( 600, 600, -10, 10, -10, 10,-1,-1, (1/4)(x^2+10x+25))) }}}