Question 193521
{{{x^2+8y+4x-4=0}}} Start with the given equation.



{{{x^2+4x+8y-4=0}}} Rearrange the terms.



Now we need to complete the square for the "x" terms.



Take half of the x coefficient 4 to get 2. Square 2 to get 4.



{{{x^2+4x+4+8y-4=4}}} Add the previous result 4 to both sides.



{{{x^2+4x+4+8y-4-4=0}}} Subtract 4 from both sides.



{{{x^2+4x+4+8y-8=0}}} Combine like terms.



{{{(x+2)^2+8y-8=0}}} Factor {{{x^2+4x+4}}} to get {{{(x+2)^2}}}



{{{8y-8=-(x+2)^2}}} Subtract {{{(x+2)^2}}} from both sides



{{{8(y-1)=-(x+2)^2}}} Factor out the GCF 8 from the left side



{{{y-1=-(1/8)(x+2)^2}}} Multiply both sides by {{{1/8}}}



{{{y-1=-(1/8)(x-(-2))^2}}} Rewrite {{{x+2}}} as {{{x-(-2)}}}




Now the equation {{{y-1=-(1/8)(x-(-2))^2}}} is in the form {{{y-k=a(x-h)^2}}} where {{{a=-1/8}}}, {{{h=-2}}} and {{{k=1}}}



Vertex:


The vertex of {{{y-k=a(x-h)^2}}} is (h,k). Since {{{h=-2}}} and {{{k=1}}}, the vertex of {{{y-1=-(1/8)(x-(-2))^2}}} is (-2,1)



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Focus: 


First, we define {{{p=1/(4a)}}} where the absolute value of "p" is the distance from the vertex to the focus. 


Note: the formula {{{p=1/(4a)}}} is derived from the equation {{{(x-h)^2=4p(y-k)}}}



The focus of {{{y-k=a(x-h)^2}}} is the point *[Tex \LARGE \left(h,k+p\right)]. In other words, simply add the value of "p" to the y coordinate of the vertex to get the focus.



So let's find "p": {{{p=1/(4a)=1/(4(-1/8))=1/(-1/2)=-2}}}. So {{{p=-2}}}



Since {{{h=-2}}}, {{{k=1}}}, and {{{p=-2}}}, this means that {{{k+p=1-2=-1}}}



So the focus of the form *[Tex \LARGE \left(h,k+p\right)] is the point (-2,-1)



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Directrix:




The directrix of {{{y-k=a(x-h)^2}}} is the equation {{{y=k-p}}}. 



Since {{{y=k-p=1-(-2)=3}}}, this means that the directrix is {{{y=3}}}



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Axis of Symmetry:



In this case, the axis of symmetry is simply the equation of the vertical line through the vertex.



Since {{{h=-2}}}, this means that the axis of symmetry is {{{x=-2}}}




Visual Check:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(-1/8)(x+2)^2+1),
green(line(-20,3,20,3)),
blue(line(-2,-20,-2,20)),
circle(-2,1,0.08),circle(-2,1,0.10),circle(-2,1,0.12),
circle(-2,-1,0.08),circle(-2,-1,0.10),circle(-2,-1,0.12)
)}}}



Graph of {{{x^2+8y+4x-4=0}}} with the vertex, focus, directrix (green) and axis of symmetry (blue)