SOLUTION: write the equation for the graph with vertex (-2,-7) and focus (1,-7). a. (x+2)^2=12(y+7) c. (y+2)^2=12(x+7) b. (y+7)^2=12(x+2) d. (y+7)^2=-12(x+2)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: write the equation for the graph with vertex (-2,-7) and focus (1,-7). a. (x+2)^2=12(y+7) c. (y+2)^2=12(x+7) b. (y+7)^2=12(x+2) d. (y+7)^2=-12(x+2)      Log On


   



Question 698405: write the equation for the graph with vertex (-2,-7) and focus (1,-7).
a. (x+2)^2=12(y+7) c. (y+2)^2=12(x+7)
b. (y+7)^2=12(x+2) d. (y+7)^2=-12(x+2)

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
write the equation for the graph with vertex (-2,-7) and focus (1,-7)
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This is a parabola that opens right.
Its standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
given vertex: (-2,-7)
axis of symmetry: y=-7
p=3 (-2 to 1) (distance from vertex to focus on the axis of symmetry)
4p=12
equation: (y+7)^2=12(x+2)