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Question 614352: Graph the parabola by finding the vertex, directrix, and opening. Equation given 4x-3=-ysq.+4y+1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Graph the parabola by finding the vertex, directrix, and opening. Equation given
4x-3=-ysq.+4y+1
4x-3=-y^2+4y+1
complete the square
4x-3=-(y^2-4y+4)+1+4
4x-3=-(y-2)^2+5
4x-8=-(y-2)^2
-(y-2)^2=4x-8
(y-2)^2=-4(x-2)
This is an equation of a parabola that opens leftwards.
Its standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation: (y-2)^2=-4(x-2)
Vertex: (2,2)
axis of symmetry: y=2
4p=4
p=1
directrix: x=3 (one unit to the right of the vertex on the axis of symmetry)
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