SOLUTION: write the equation for the graph with vertex (-6,3) and focus (-6,0.25). a. (x+6)^2=11(y-3) c. (y-3)^2=-11(x+6) b. (x+6)^2=-11(y-3) d. (y+6)^2=-11(x-3)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: write the equation for the graph with vertex (-6,3) and focus (-6,0.25). a. (x+6)^2=11(y-3) c. (y-3)^2=-11(x+6) b. (x+6)^2=-11(y-3) d. (y+6)^2=-11(x-3)      Log On


   



Question 698407: write the equation for the graph with vertex (-6,3) and focus (-6,0.25).
a. (x+6)^2=11(y-3) c. (y-3)^2=-11(x+6)
b. (x+6)^2=-11(y-3) d. (y+6)^2=-11(x-3)

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
write the equation for the graph with vertex (-6,3) and focus (-6,0.25).
This is a parabola that opens downwards.
Its standard form of equation: (x-h)^2=-4p(y-k), (h,k)=coordinates of vertex
Given vertex: (-6,3)
Axis of symmetry: x=-6
given focus:(-6,0.25)
p=3-.25=2.75 (distance from vertex to focus on the axis of symmetry)
4p=11
Equation: (x+6)^2=-11(y-3) (ans. d)