Question 1042680
Standard is ax^2+bx+c=0
Start with x^2-2x-6=0
The vertex x value is -b/2a
Here, that would be -(-2)/2(1)=2/2=1
So the x-value is 1
Plug that into the equation and the y value is 1^2-(2)(1)-6=1-2-6=-7
The vertex is (1,-7)
The vertex form is y-=a(x-h)+k.  Change the sign of the x coordinate and keep the sign of the y.
y=(x-1)^2-7
You can always check by foiling out the square term.
It is x^2-2x+1-7=x^2-2x-6
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3x^2-6x-5=0
divide the first two terms by 3 and put the 3 out front.
3(x^2-2x)-5=0
complete the square by taking half the x term and squaring it.  Whatever it is, subtract it from the constant to keep the equation balanced. We have added a 1 inside the parentheses and the whole inside term is multiplied by 3. We have to subtract 3 from the equation, too.
3(x^2-2x+1)-5-3=3(x^2-2x+1)-8
Now write it as the square.
3(x-1)^2-8
The vertex is at (1,-8), change the sign of the number after the x and keep the sign of the constant at the end.
{{{graph(300,300,-10,10,-10,10,3x^2-6x-5)}}}