Question 208165
{{{x^2-2x-8}}}
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Complete the square and convert to "vertex form" or "turning point":
y= a(x-h)^2+k 
where (h,k) is the turning point (vertex)
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{{{x^2-2x-8}}}
{{{(x^2-2x+1)-8-1}}}
{{{(x-1)(x-1)-9}}}
{{{(x-1)^2-9}}}
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Therefore, the vertex is at (1,-9)