we have that
[tex]\frac{dy}{dx}=kx-6[/tex]At (2,1) the curve has a turning point ----> that means ---> the derivative is equal to zero
so
[tex]0=kx-6[/tex]substitute the x-coordinate of point (2,1)
x=2
[tex]\begin{gathered} 0=k(2)-6 \\ 2k=6 \\ k=3 \end{gathered}[/tex]Part a)
the valuie of k=3
Part b
the equation of the curve
we have
[tex]\frac{dy}{dx}=3x-6[/tex]therefore
[tex]y=\int ^{}_{}(3x-6)dx[/tex][tex]y=\frac{3x^2}{2}-6x+C[/tex]