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