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Vertex form of y=(x-2)(x+2)​

1 Answer

8 votes

Answer:


\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)

User RonanC
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