Question 1045379
Let the vertex be {{{V(k,k)}}} .
The equation of a parabola with that vertex and axis parallel to the x-axis is
{{{x-k=a(x-k)^2}}} .
Substituting the coordinates of the given points, we have
{{{system(x=6,y=-2)}}}--->{{{6-k=a(-2-k)^2}}}
{{{system(x=3,y=4)}}}--->{{{3-k=a(4-k)^2}}}
Those two equations form a system of equation we have to solve to find {{{a}}} and {{{k}}} .
Subtracting one equation from the other, we have
{{{6-k-(3-k)=a(-2-k)^2-a(4-k)^2}}}
{{{6-k-3+k)=a((-2-k)^2-(4-k)^2)}}}
{{{3=a(-2-k+(4-k))(-2-k-(4-k))}}}
{{{3=a(-2-k+4-k)(-2-k-4+k)}}}
{{{3=a(-2k+2)(-6)}}}
{{{3=a(-2)(k-1)(-6)}}}
{{{3=12a(k-1)}}}
{{{3/12(k-1)=a}}}
{{{1/4(k-1)=a}}}
Substituting the expression {{{1/4(k-1)}}} for {{{a}}} in {{{3-k=a(4-k)^2}}} we get
{{{3-k=(4-k)^2/4(k-1)}}}-->{{{4(k-1)(3-k)=(4-k)^2}}}-->
{{{4(-k^2+4k-3)=(4-k)^2}}}-->{{{-4k^2+16k-12=16-8k+k^2}}}
-->{{{0=16-16k+k^2+4k^2-16k+12}}}-->{{{0=5k^2-24k+28}}}
The solutions to that equation are {{{k=2}}} and {{{k=24/5}}} .
Only one of those solutions should work, because
through any 3 points, (such as (6,-2) , (3,4), and {{{V(2,2)}}} )
passes only one parabola with axis parallel to the x-axis.
Substituting {{{2}}} for {{{k}}} in the original equation
{{{3-k=a(4-k)^2}}} , we get
{{{3-2=a(4-2)^2}}}--->{{{1=4a}}}--->{{{a=1/4}}}
The values {{{system(k=2,a=1/4)}}} also satisfy equation {{{6-k=a(-2-k)^2}}} .
So, {{{highlight(x-2=(y-2)^2/4)}}} is the equation of a parabola with vertex on the line {{{y=x}}} , axis parallel to the x-axis, and passing through (6,-2) and (3,4).
The equation can be written in other, equivalent forms.
{{{highlight(x-2=(y-2)^2/4)}}}<-->{{{highlight(x=(y-2)^2/4+2)}}}<-->{{{highlight(x=(1/4)y^2-x+3)}}}
 
NOTE: The solution {{{k=14/5}}} does not yield the same a value when substituted in the two original equations.