SOLUTION: The graph of a parabola passes through the points (3/2, 4/3) and (0, −6) and has a horizontal tangent line at (3/2, 4/3). Find an equation for the parabola and sketch its graph

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Question 1153918: The graph of a parabola passes through the points (3/2, 4/3) and (0, −6) and
has a horizontal tangent line at (3/2, 4/3). Find an equation for the parabola
and sketch its graph.

Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

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) = ,


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    = -6,   or


     = -6.


It gives


     =  =  = ,


hence


    a =  = .


Thus the equation of the parabola is  


    y(x) = .


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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