Question 860380
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^2+cy+4}}}
{{{x=-1(y^2-cy-4)}}}
The term to use for completing the square is {{{(c/2)^2}}}.
{{{x=-1(y^2-cy+(c/2)^2-4-(c/2)^2)}}}
{{{x=-1((y-c/2)^2-(4+(c/2)^2))}}}
{{{x=-1((y-c/2)^2-(16/4+c^2/4))}}}
{{{x=-1((y-c/2)^2-(c^2+16)/4)}}}
{{{highlight(x=-1*(y-c/2)^2+(c^2+16)/4)}}}


The axis of symmetry is horizontal, and will be on some line {{{y=(c^2+16)/4}}}.  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/2}}}.  The vertex point is at {{{x=(c^2+16)/4}}}, {{{y=c/2}}}.


Note that c may be positive or negative or zero.