Question 1062373
The general form {{{x=ay^2+by+c}}} and the two given points should allow you to make a linear system of two equations in three unknown variables.  You should be able to eliminate two of them.  Your next equation y=x might allow you to find the value for the last unknown variable.  Converting from general form into standard form might also help.


The best I have been able to find so far is {{{system(4a+2b=-1,x=a(y-(b/2a))^2+c-b^2/4a)}}}, 
and the vertex being on line y=x, would be some point  ( {{{c-b^2/(4a)}}}, {{{b/(2a)}}} ).


The linear equation of a and b, and the expected vertex being on line y=x justify a system {{{system(c-b^2/(4a)=b/2a,4a+2b=-1,4a-2b+c=6,16a+4b+c=3)}}}.  Still not a final answer.


The system most likely to give something meaningful might be  the two specific point equations, and the vertex coordinate relationship equation:
{{{system(4a-2b+c=6,16a+4b+c=3,4ac-b^2=2b)}}}.


Solve the first of those for c and substitute into the next two equations.  This should be able to make a system
{{{system(4a+2b=-1,16a^2+b^2-24a-8ab+2b=0)}}}.
Notice that the first equation here has a term  {{{4a}}}, and that a few of the other equation's terms can be factored with
{{{4a}}, so maybe a substitution for {{{4a}}} might be useful.  This could give a single equation in just the one variable, b.


Do that and you get {{{9b^2+21b+7=0}}}.  Quadratic formula solution will then after simplification give you just for b,
{{{highlight(b=(-7+- sqrt(21))/6)}}}.


Still a ways to go to get the possible "a" and "c" values.