You can either transform this quadratic equation into a vertex form which would rather painful or use derivatives.
I'll use dervatives. We know that vertex of a quadratic function is either its maxima or its minima, since the leading coefficient [tex]-x^2[/tex] has a negative prefix that means we get a downward turned parabola with minima being the vertex.
First we take the derivative with respect to x,
[tex]\dfrac{d}{dx}-x^2-2x+3=-2x-2[/tex]
The derivative is esentially information what is the slope of a function at a particular x. When the slope is 0 we reached some sort of turning point, such as minima.
We therefore do,
[tex]-2x-2=0\implies x=-1[/tex]
So at [tex]x=-1[/tex] there appears to be a minima or x-coordinate of the vertex of the function.
Plug the coordinate into the function to get y,
[tex]y=f(-1)=-1+2+3=4[/tex]
So the vertex of the function is at [tex]\boxed{(-1,4)}[/tex].
Assuming you don't know derivatives, there is another way.
First compute the roots of the function,
[tex]-x^2-2x+3=0[/tex]
[tex]-(x-1)(x+3)=0[/tex]
[tex]x_1=1,x_2=-3[/tex]
In the middle between [tex]x_1,x_2[/tex] is an x coordinate of a vertex,
[tex]x=\dfrac{x_1+x_2}{2}=\dfrac{1-3}{2}=-1[/tex]
Just like we had before, we compute for y, [tex]h(-1)=4[/tex] and again the result is [tex]\boxed{(-1,4)}[/tex].
Hope this helps :)
