Question 898531
<pre>
I'll just do the second one.  You do the others the same way:

{{{y}}}{{{""=""}}}{{{3x^2-4x+5}}}

Factor 3 out of the first two terms only.  Since 4 is not 
divisible by 3 and we have to take out a 3 anyway, we must 
divide the 4 by 3 and get a fraction. 

{{{y}}}{{{""=""}}}{{{3(x^2-expr(4/3)x)+5}}}

Next we complete the square inside the parentheses:

1. Multiply the coefficient of the term in x, which is {{{-4/3}}},
   by {{{1/2}}}, getting {{{expr(-4/3)*expr(1/2)=-4/6=-2/3}}}
2. Square this {{{(-2/3)^2=4/9}}}
3. Add and subtract {{{4/9}}} inside the parentheses:

{{{y}}}{{{""=""}}}{{{3(x[""]^2-expr(4/3)x+4/9-4/9)+5}}}

Group the first three terms in smaller parentheses inside the
larger parentheses:

{{{y}}}{{{""=""}}}{{{3((x^2-expr(4/3)x+4/9)-expr(4/9)[""])+5}}}

The inner parentheses trinomial factors like this:{{{x^2-expr(4/3)x+4/9=(x-2/3)(x-2/3)=(x-2/3)^2}}}

{{{y}}}{{{""=""}}}{{{3((x-2/3)^2[""]-4/9)+5}}}
 
Now we distribute to remove the large parentheses, leaving the smaller
parentheses intact:

{{{y}}}{{{""=""}}}{{{3(x-2/3)^2-3*expr(4/9)+5}}}

Then we complete the fraction work on the end.
Cancel the 3 into the 9 and we have

{{{y}}}{{{""=""}}}{{{3(x-2/3)^2-4/3+5}}}

Get a common denominator of 3 by multiplying the 5 by {{{3/3}}}

{{{y}}}{{{""=""}}}{{{3(x-2/3)^2-4/3+15/3}}}

And the final vertex form is:

{{{y}}}{{{""=""}}}{{{3(x-2/3)^2+11/3}}}

Edwin</pre>