Vertex of this parabola?

Question

x= y^2 + 2y + 5

Answer 1

Hi,

x= y^2 + 2y + 5

x= y² + 2y + 1 + 4

x= (y + 1)² + 4

The vertex is at (4,-1). <==ANSWER

I hope that helps!! :-)

Answer 2

(Note:  The parabola has y^2 instead of x^2.

It opens to the right and not up.)

If y = f(x) = ax^2 + bx + c, then the vertex is at

(- b/2a, f(- b/2a) ).

If x = f(y) = ay^2 + by + c, then the vertex is at

( f( -b/2a), - b/2a).

  

Answer 3

find derivative of polynomial function

derivative= 2y + 2

2y+2=0 | y=-1

x=(-1)^2 +2(-1)+5

x=4

(4,-1)