Question 1133782
focus, latus rectum, directrix, axis of symmetry

{{{-2(x+3)=(y-1)^2 }}}

{{{4p(x-h)=(y-k)^2}}} is the standard equation for a right-left facing parabola with vertex at ({{{h}}},{{{k}}})  and a focal length {{{abs(p)}}}

Rewrite{{{-2(x+3)=(y-1)^2 }}} in the standard form

as you can see {{{4p=-2}}}=>{{{p=-1/2}}}

{{{ 4(-1/2)(x-(-3))=(y-1)^2}}}

Therefore parabola properties are: 

({{{h}}},{{{k}}}) =({{{-3}}},{{{1}}}) 
{{{p=-1/2}}}

Parabola is symmetric around the x-axis and so the focus lies a distance {{{p}}} from the center ({{{-3+p}}}, {{{1}}} ) along the x-axis 

({{{-3-1/2}}}, {{{1}}} )....plug in {{{p=-1/2}}}

focus: ({{{-7/2}}}, {{{1}}} )

directrix is 
{{{x=-3-p}}}
{{{x = -3-(-1/2)}}}
{{{x = -6/2+1/2}}}
{{{x = -5/2}}}


latus rectum:
{{{-2(x+3)=(y-1)^2 }}}............switch sides

{{{(y-1)^2=-2(x+3) }}}......solve for {{{y}}}

{{{y-1= sqrt(-2(x+3) )}}}

{{{y= sqrt(-2(x+3) )+1}}}

{{{y= sqrt(2(-x-3) )+1}}}

=> solutions:

{{{y= sqrt(2)sqrt(-x-3) +1}}}

{{{y= 1-sqrt(2)sqrt(-x-3) }}}


Axis of symmetry is a line parallel to the x -axis which intersects the vertex: 
{{{y=1}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-3,1,.12),locate(-3,1,V(-3,1)),line(-5/2,10,-5/2,-10),circle(-7/2,1,.12),locate(-7/2,1,f(-7/2,1)),
graph( 600, 600, -10, 10, -10, 10, sqrt(2(-x-3) )+1, 1-sqrt(2)sqrt(-x-3),1,1)) }}}