Answer:
[tex]\mathrm{Minimum}\:\left(0,\:-4\right)[/tex]
Step-by-step explanation:
[tex]y=\left(x-2\right)\left(x+2\right)[/tex]
[tex]Parabola\:equation\:in\:vertex\:form:[/tex] [tex]y=a\left(x-h\right)^2+k\:\mathrm{\:is\:the\:equation\:in\:vertex\:form\:for\:an\:up-down\:facing\:parabola\:with\:vertex\:at}[/tex] [tex](h,k)[/tex]
Rewrite y=(x-2)(x+2) in vertex form:
[tex]y=\left(x-0\right)^2-4[/tex]
[tex]\mathrm{The\:parabola\:params\:are:}[/tex]
[tex]a=1,\:h=0,\:k=-4[/tex]
[tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}[/tex]
[tex]\left(h,\:k\right)=\left(0,\:-4\right)[/tex]
[tex]\mathrm{If}\:a < 0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}[/tex]
[tex]\mathrm{If}\:a > 0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}[/tex]
[tex]a=1[/tex]
[tex]\mathrm{Minimum}\:\left(0,\:-4\right)[/tex]
