Question 1153918
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Since the parabola has a horizontal tangent line at the point (3/2,4/3), it means 

that this point (3/2,4/3) is the vertex of the parabola.


Hence, the parabola has the form


    y(x) = {{{a*(x-3/2)^2 + 4/3}}},


where "a" is a coefficient (real number), now unknown.


To find "a", use the fact that the plot of the parabola passes through the point (0,-6). It means that


    y(0) = -6,   or   {{{a*(0 - 3/2)^2 + 4/3}}} = -6,   or


    {{{a*(3/2)^2 + 4/3}}} = -6.


It gives


    {{{a*(9/4)}}} = {{{-6 - 4/3}}} = {{{(-18-4)/3}}} = {{{-22/3}}},


hence


    a = {{{(4/9)*(-22/3)}}} = {{{-88/27}}}.


Thus the equation of the parabola is  


    y(x) = {{{(-88/27)*(x-3/2)^2 + 4/3}}}.


You can transform it to any other equivalent form you wish.


Regarding the plot, do it on your own.


You may use free of charge plotting tools from the Internet.
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