Question 968946
Some form of {{{x=a(y-k)^2}}}, because you are not sure where along the y-axis is the intercept.


Your two given points will allow to find a and k.



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The expected format equation, {{{x=a(y-k)^2}}} and the two points to be on the graph:


{{{system(2=a(4-k)^2,8=a(-2-k)^2)}}}, so this system is solvable for a and k.
Use any whatever good algebra skills you have.


{{{system(2/(4-k)^2=a,8/(-2-k)^2=a)}}}


{{{2/(4-k)^2=8/(-2-k)^2}}}


their reciprocals are therefore also equal.
{{{(4-k)^2/2=(-2-k)^2/8}}}


{{{8(4-k)^2=2(-2-k)^2}}}


{{{4(4-k)^2=(-1)^2(2+k)^2}}}


{{{4(4-k)^2=(2+k)^2}}}


{{{4(16-8k+k^2)=4+4k+k^2}}}


{{{64-32k+4k^2=4+4k+k^2}}}


{{{3k^2-36k+60=0}}}


{{{k^2-12k+20=0}}}

{{{(k-2)(k-10)=0}}}


The possibility of two different values for k, and each of them will have a corresponding value for a.


{{{highlight(k=2)}}}  OR  {{{highlight(k=10)}}}.


You continue on to find values of "a".