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Question 64382: please help me find the vertex, axis of symm., x and y intercepts, focus and directrix. It would be helpful if you could sketch the graphs showing the points.
y=-1/4(x-2)^2+4 and
x=3(y+1)^2-2
Thank you.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! please help me find the vertex, axis of symm., x and y intercepts, focus and directrix. It would be helpful if you could sketch the graphs showing the points.
y=-1/4(x-2)^2+4
y-4=(-1/4)(x-2)^2
Vertex: (2,4)
Axis of sym: x=2
x-int: let y=0 then (1/4)(x-2)^2=4; (x-2)^2=16; x-2=4 or x-2=-4; x=6 or -2
y-int: let x=0 then y=(-1/4)(-2)^+4=-1+4=3
Note: 4p=-(1/4) so p=-1/16
Focus = (2,4+(-1/6))=(2,15/16)
Directrix: y=4+(1/16)=17/16
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and
x=3(y+1)^2-2
x+2=3(y+1)^2
Vertex: (-2,-1)
Axis of sym: y=-1
x-int: let y=0 then x=1
y-int: let x=0 then (y+1)^2=2/3 ; y=-1+sqrt(2/3) or y=-1-sqrt(2/3)
Note: 4p=3 so p=3/4
Focus: (-2+(3/4),-1)=(-5/4,-1)
Directrix: x=-2-(3/4)=-11/4
This site software cannot graph this as it is not a function.
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Cheers,
Stan H.
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