Question 866690
That would not be a standard equation for y for a parabola.  How about standard form for x in terms of y instead?  Your given equation has the square for y already completed and only x in the power of 1, so standard form as x in terms of y is possible.


You can solve your equation for y in terms of x, but it will not be a standard form parabola equation.
{{{(y-4)^2=2(x+11)}}}
{{{y-4=0+- sqrt(2(x+11))}}}
{{{y=4+- sqrt(2(x+11))}}}
{{{highlight(y=4+- sqrt(2)sqrt(x+11))}}}, two square root functions, each being half a parabola.  Each is stretched horizontally, and vertex is (-11,4); axis of symmetry between the two functions is y=4.