SOLUTION: How do you find the Vertex, focus, directric of the parabola and sketch its graph for the below equation? y=2x^2 Vertex: ______ Focus: _____ Directrix _______ Axis of Symmet

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do you find the Vertex, focus, directric of the parabola and sketch its graph for the below equation? y=2x^2 Vertex: ______ Focus: _____ Directrix _______ Axis of Symmet      Log On


   



Question 970692: How do you find the Vertex, focus, directric of the parabola and sketch its graph for the below equation?
y=2x^2
Vertex: ______
Focus: _____
Directrix _______
Axis of Symmetry: Vertical or Horizontal

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How do you find the Vertex, focus, directrix of the parabola? Sketch its graph for the below equation?
y=2x^2
Vertex: ______
vertex occurs where x = -b/(2a) = -0/4 = 0
then f(0) = 0
So the vertex is (0,0)
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Focus: _____
Write the equation is vertex form::
(x-0)^2 = (1/2)(y-0)
Then 4p = 1/2, so p = 1/8
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Focus is at (0,0+p) = (0,1/8)
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Directrix _______
Directrix is y = 0-p = -1/8
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Axis of Symmetry: Vertical or Horizontal
The vertical axis of symmetry is x = 0
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Cheers,
Stan H.
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