SOLUTION: Find the vertex, focus, directrix, and axis of symmetry of each parabola. x^2+8y+4x-4=0 We have been using the formulas y-k=a(x-h)^2 and x-h=a(y-k)^2 I don't understand which to

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, focus, directrix, and axis of symmetry of each parabola. x^2+8y+4x-4=0 We have been using the formulas y-k=a(x-h)^2 and x-h=a(y-k)^2 I don't understand which to      Log On


   

Question 193531This question is from textbook algebra and trigonometry structure and method book 2
: Find the vertex, focus, directrix, and axis of symmetry of each parabola.
x^2+8y+4x-4=0
We have been using the formulas y-k=a(x-h)^2 and x-h=a(y-k)^2 I don't understand which to use here, even after completing the square and its confusing because I don't know which is which. please explain and be specific. thank you. This question is from textbook algebra and trigonometry structure and method book 2

Found 2 solutions by Mathtut, jim_thompson5910:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
8y=-x%5E2-4x%2B4
:
8y=-1%28x%5E2%2B4x%2B4%29%2B4-4
:
8y=-1%28x%2B2%29%5E2
:
y=%28-1%2F8%29%28x%2B2%29%5E2
:
a=coefficient of x%5E2 term so a=-1/8
:
so vertex is (-2,0)
focus is (-2,0+1/4a)-->1/4a=1/(4(-1/8)=1/(-1/2)=-2--> (-2,-2)
:
directrix is at y=-1/4a=-1/4(-1/8)=-1/(-1/2)=2 so y=2
:
axis of symmetry is x=-2
:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%28-1%2F8%29%28x%2B2%29%5E2%29

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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