Question 1063585
{{{x=-4y^2+16}}}
{{{x=-4(y^2-4)}}}
{{{x=-4(y+2)(y-2)}}}


Intercepts:  (0,2), (0,-2), (16,0).


x is a quadratic in terms of y; the y is squared.  The coefficient in x=-4(y+2)(y-2) or x=-4y^2+16 on the leading term is negative.  Parabola opens to the left; symmetry axis is parallel to the horizontal axis.



{{{graph(300,300,-2,18,-10,10,-sqrt(-x/4+4),sqrt(-x/4+4))}}}






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4y^2=-x+16
y^2=-x/4+4
y=0+- sqrt(-x/4+4)