SOLUTION: Sketch the graph of the quadratic equation x+y^2-cy=4, where c is a positive constant. Since y is squared and x is not, would it be a parabola that opens left/right? How do I f

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Sketch the graph of the quadratic equation x+y^2-cy=4, where c is a positive constant. Since y is squared and x is not, would it be a parabola that opens left/right? How do I f      Log On


   

Question 860380: Sketch the graph of the quadratic equation x+y^2-cy=4, where c is a positive constant.
Since y is squared and x is not, would it be a parabola that opens left/right?
How do I find the vertex and focus if it has a c?
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The equation would be a parabola which opens left or right. Complete The Square for y and put the equation into standard form. x= something.

x=-y%5E2%2Bcy%2B4
x=-1%28y%5E2-cy-4%29
The term to use for completing the square is %28c%2F2%29%5E2.
x=-1%28y%5E2-cy%2B%28c%2F2%29%5E2-4-%28c%2F2%29%5E2%29
x=-1%28%28y-c%2F2%29%5E2-%284%2B%28c%2F2%29%5E2%29%29
x=-1%28%28y-c%2F2%29%5E2-%2816%2F4%2Bc%5E2%2F4%29%29
x=-1%28%28y-c%2F2%29%5E2-%28c%5E2%2B16%29%2F4%29
highlight%28x=-1%2A%28y-c%2F2%29%5E2%2B%28c%5E2%2B16%29%2F4%29

The axis of symmetry is horizontal, and will be on some line y=%28c%5E2%2B16%29%2F4. The graph opens to the left. (The negative 1 coefficient says, "open to the left".) The farthest to the right that any points are for the graph is at c%2F2. The vertex point is at x=%28c%5E2%2B16%29%2F4, y=c%2F2.

Note that c may be positive or negative or zero.