Question 889601
{{{highlight_green(x=a(y-k)^2+h)}}}, parabola parallel to the horizontal axis.


The line y=x containing the vertex means that the two coordinates of this point are equal in value.
Call this vertex (n,n).
The equation of the parabola can be expressed, {{{highlight_green(x=a(y-n)^2+n)}}}.


The variables, once the given points are substituted, will be a and n.  General form should give
a system of equations which we could solve for a and n.


{{{x=a(y^2-2yn+n^2)+n}}}
{{{x=ay^2-2ayn+an^2+n}}}
{{{ay^2-2ayn+an^2+n-x=0}}}
-
Maybe alternatively
{{{a(y^2-2yn+n^2)=x-n}}}
{{{a=(x-n)/(y^2-2yn+n^2)}}}, because this gives a on one side and an expression in terms of n on the other side.
Knowing that a and n are constants, any point on the parabola should give equal expressions for the member on
the right-side.


Using the two given points, those points (6,-2) and (3,4),
{{{a=(6-n)/((-2)^2-2(-2)n+n^2)}}}
and
{{{a=(3-n)/(4^2-2*4+n^2)}}}


The two expressions for a must be equal.
{{{(6-n)/((-2)^2-2(-2)n+n^2)=(3-n)/(4^2-2*4+n^2)}}}
{{{(6-n)/(4+4n+n^2)=(3-n)/(16-8n+n^2)}}}
{{{(6-n)(16-8n+n^2)=(3-n)(4+4n+n^2)}}}
{{{96-48n+6n^2-16n+8n^2-n^3=12+12n+3n^2-4n-4n^2-n^3}}}
{{{96-48n-16n+6n^2+8n^2=12+12n-4n+3n^2-4n^2}}}
{{{96-64n+14n^2=12+8n-n^2}}}
{{{96-12-64n-8n+14n^2+n^2=0}}}
{{{15n^2-72n+84=0}}}, Divide this by 3;
{{{highlight_green(5n^2-24n+28=0)}}}


How is the discriminant?
{{{24^2-4*5*28}}}
{{{576-20*28}}}
{{{576-560}}}
{{{highlight_green(16)}}}, a perfect square whole number.


Now, what can be n?
{{{n=(24+- sqrt(16))/(2*5)}}}
{{{n=(24+- 4)/10}}}
Either {{{highlight(n=2)}}}  Or  {{{highlight(n=14/5)}}}


... Still not finished.  We want to use these values for n to get our value for a; probably a single (or maybe not) value.  You will have to try each of the solutions for n and find out.