How do you find the maximum/ minimum of the quadratic equation: f(x)=x^2+2x+4

If you only know algebar, find the vertex
we see it opens up
so minimum
so the x value of the vertex is -b/2a for ax^2+bx+c=y

so
f(x)=1x^2+2x+4
x value of vertex is -2/(2*1)=-2/2=-1

x value of the minimum is x=-1
if we find f(-1) we get 3

minimum value is 3 at x=-1