Verified Answer :

\(y=ax^2+bx+c\\\\vertex:(x_v;\ y_v)\\\\x_v=\frac{-b}{2a}\ and\ y_v=f(x_v)\)


\(y=x^2-2x-1\\\\a=1;\ b=-2;\ c=-1\\\\x_v=\frac{-(-2)}{2\cdot1}=\frac{2}{2}=1\\\\y_v=1^2-2\cdot1-1=1-2-1=-2\\\\Answer:(1;-2)\)

konrad509

\(y=a(x-h)^2+k \Rightarrow \hbox{vertex}=(h,k)\\\\ y=x^2-2x-1\\ y=x^2-2x+1-2\\ y=(x-1)^2-2 \Rightarrow \hbox{vertex}=(1,-2)\)

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