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Posted by on Monday, December 5, 2011 at 10:06pm.

How do i graph this hyperbola?
X^2+6x-3y^2=0

  • college algebra (math 90) - , Monday, December 5, 2011 at 10:59pm

    complete the square
    x^2 + 6x + 9 - 3y^2 = 9
    (x+3)^2 - 3y^2 = 9
    divide by 9
    (x+3)^2 /9 - y^2/3 = 1

    centre (-3,0)
    a = 3, b = √3

    vertices at (3,0) and (-3,0)
    faintly sketch a rectangle with corners
    at (±3,±√3)
    draw in the diagonals and extend them
    sketch the hyperbola with the diagonals as asymptotes and the vertices as noted above

  • college algebra (math 90) - , Tuesday, December 6, 2011 at 10:27am

    except the whole thing is shifted 3 units to the left - because it's (x+3)^2

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