SOLUTION: A conic section has the equation x^2-6x-y+4y-4=0. Which of the following are true? Check all that apply A. The equation is a parabola with its vertex at (3, 2). B.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A conic section has the equation x^2-6x-y+4y-4=0. Which of the following are true? Check all that apply A. The equation is a parabola with its vertex at (3, 2). B.       Log On


   

Question 467197: A conic section has the equation x^2-6x-y+4y-4=0. Which of the following are true? Check all that apply


A. The equation is a parabola with its vertex at (3, 2).

B. The equation is a circle with its center at (3, 2).

C. The equation is a hyperbola with one vertex at (0, 2).

D. The radius is 1.

E. The radius is 2.

F. The graph has asymptotes at y=x-1 and y=-x+5
Found 2 solutions by Big Poop, solver91311:
Answer by Big Poop(157) About Me  (Show Source):
Answer by solver91311(24713) About Me  (Show Source):
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The equation of a circle has both and terms with equal coefficients.

The equation of a hyperbola has both and terms with coefficients that have opposite signs.

If the graph has a radius, then it is a circle.

Polynomial functions do not have asymptotes.

That only leaves one possibility.

John

My calculator said it, I believe it, that settles it
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