Vertex, y-intercept, x-intercept of this parabola?

Question

x= y^2+2y+5

The teacher mentioned to not confuse it as y=x^2+2x+5

Answer 1

Answer it the same way as you would y=x^2+2x+5

plug in y=0 to solve for the x-intercept.

plug in x=0 to solve for the y-intercept.

Take the derivative of y^2+2y+5 with respect to y and solve for 0 to find the vortex.

Answer 2

im using calculus for this but the vertex is 4,-1 the x intercept is 5,0 and it does not cross the y intercept at all