Question

Vertex form of y=(x-2)(x+2)​

Answer 1

`\mathrm{Minimum}\:\left(0,\:-4\right)`

Explanation:

`y=\left(x-2\right)\left(x+2\right)`

`Parabola\:equation\:in\:vertex\:form:`
`y=a\left(x-h\right)^2+k\:\mathrm{\:is\:the\:equation\:in\:vertex\:form\:for\:an\:up-down\:facing\:parabola\:with\:vertex\:at}`
`(h,k)`

Rewrite y=(x-2)(x+2) in vertex form:

`y=\left(x-0\right)^2-4`

`\mathrm{The\:parabola\:params\:are:}`

`a=1,\:h=0,\:k=-4`

`\mathrm{Therefore\:the\:parabola\:vertex\:is}`

`\left(h,\:k\right)=\left(0,\:-4\right)`

`\mathrm{If}\:a < 0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}`

`\mathrm{If}\:a > 0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}`

`a=1`

`\mathrm{Minimum}\:\left(0,\:-4\right)`