[tex]f(x)=x^2+4x-12=x^2+2x\cdot2+2^2-2^2-12=(x+2)^2-4-12=(x+2)^2-16[/tex]
The coordinates of the vertex (-2; -16).
[tex]f(0)=0^2+4\cdot0-12=-12\to(0;-12)[/tex]
The zeros:
[tex]f(x)=0\Rightarrow (x+2)^2-16=0\ \ \ |+16\\\\(x+2)^2=16\to x+2=\pm\sqrt{16}\\\\x+2=-4\ \vee\ x+2=4\ \ \ |-2\\\\x=-6\ \vee\ x=2\\\\(-6;\ 0);\ (2;\ 0)[/tex]
