SOLUTION: We are finding the vertex, focus or directive. Then finding an equation and graphing.
the problem I am having trouble with is
vertex= 0,0
Focus =3,0
so y=-3
here is the formu
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: We are finding the vertex, focus or directive. Then finding an equation and graphing.
the problem I am having trouble with is
vertex= 0,0
Focus =3,0
so y=-3
here is the formu
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Question 583308: We are finding the vertex, focus or directive. Then finding an equation and graphing.
the problem I am having trouble with is
vertex= 0,0
Focus =3,0
so y=-3
here is the formula I get
(y+3)^2=(x-3)^2+(y-0)^2
y^2+6y+9=(x-3)^2+y^2
ans I get is:
y+3/2=1/6(x-3)^2
But the teacher gives us the answers and it is x=1/12y^2
Can someone explain how she got this answer?
Thank you! Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website! the vertex and the focus lie on the axis of symmetry, which is ___ y = 0
___ this is a side-to-side parabola, not up-and-down
