Since the quadratic model is
[tex]C(t)=8.28t^2+268.6t+3500[/tex]To find the equation of the tangent differentiate C(t)
[tex]\begin{gathered} C^{\prime}(t)=8.28(2)t^{2-1}+268.6(1)t^{1-1}+0 \\ C^{\prime}(t)=16.56t+268.6 \end{gathered}[/tex]m of the tangent is equal to C'(t)
Since we need it in 2016, then substitute x by
[tex]2016-1990=26[/tex][tex]\begin{gathered} m=16.56(26)+268.6 \\ m=699.16 \end{gathered}[/tex]Then the equation of the tangent is
[tex]y=699.16t+b[/tex]To find b substitute t by 26 in C(t) to find y, then substitute x and y in the equation of the tangent
[tex]\begin{gathered} C(26)=8.28(26)^2+268.6(26)+3500 \\ C(26)=16080.88 \end{gathered}[/tex]Substitute t by 26 and y by 16080.88 in the equation of the tangent to find b
[tex]\begin{gathered} 16080.88=699.16(26)+b \\ 16080.88=18178.16+b \\ b=16080.88-18178.16 \\ b=-2097.28 \end{gathered}[/tex]The equation of the tangent is
[tex]y=699.16t-2097.28[/tex][tex]\begin{gathered} m=699.16 \\ b=-2097.28 \end{gathered}[/tex]The estimated tuition in 2016 is 16080.88
