How do you graph #x= - (y-2)^2 + 3#?

1 Answer

Answer:

Vertex #(3, 2)# and it opens to the LEFT
with intercepts at #(-1, 0), (0, 2+sqrt3),(0, 2-sqrt3)#

Explanation:

The graph is a parabola which opens to the left
#x=-(y-2)^2+3#. When we transform the equation to its vertex form
#+-4p(x-h)=(y-k)^2#, it is noticable that there is a negative sign

#-(x-3)=(y-2)^2# that is why this parabola opens to the left.
graph{x=-(y-2)^2+3[-20,20,-10,10]}

God bless...I hope the explanation is useful.