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| Question 533794:  find the vertex, focus, and directrix of the following parabola. then draw the grapg. (x-3)^2=-5(y+1)
 Answer by lwsshak3(11628)
      (Show Source): 
You can put this solution on YOUR website! find the vertex, focus, and directrix of the following parabola. then draw the grapg. (x-3)^2=-5(y+1) **
 Given equation is that of a parabola of standard form: (x-h)^2=4p(y-k), (h,k) being the (x,y) coordinates of the vertex and parabola opens downward.
 For given equation: (x-3)^2=-5(y+1)
 vertex: (3,-1)
 4p=5
 p=5/4 (distance down from vertex to focus on axis of symmetry, x=3
 focus:(3,(-1-p))=(3,(-1-(5/4))=(3,-9/4)
 directrix: y=-1+(5/4)=1/4 (a horizontal line p units up from the vertex on the axis of symmetry
 y-intercept:
 set x=0
 3^2=-5y-5
 9+5=-5y
 y=-14/5=2.8
 This gives you two points, (0,-2.8) and (6,-2.8), in addition to the coordinates of the vertex, with which you can graph given equation
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