Question

A quadratic model C(t) for the cost of tuition at a small private college in dollars for t years since 1990 is C( t ) =

Answer 1

Since the quadratic model is

`C(t)=8.28t^2+268.6t+3500`

To find the equation of the tangent differentiate C(t)

`\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}`

m of the tangent is equal to C'(t)

Since we need it in 2016, then substitute x by

`2016-1990=26`

`\begin{gathered} m=16.56(26)+268.6 \\ m=699.16 \end{gathered}`

Then the equation of the tangent is

`y=699.16t+b`

To find b substitute t by 26 in C(t) to find y, then substitute x and y in the equation of the tangent

`\begin{gathered} C(26)=8.28(26)^2+268.6(26)+3500 \\ C(26)=16080.88 \end{gathered}`

Substitute t by 26 and y by 16080.88 in the equation of the tangent to find b

`\begin{gathered} 16080.88=699.16(26)+b \\ 16080.88=18178.16+b \\ b=16080.88-18178.16 \\ b=-2097.28 \end{gathered}`

The equation of the tangent is

`y=699.16t-2097.28`
`\begin{gathered} m=699.16 \\ b=-2097.28 \end{gathered}`

The estimated tuition in 2016 is 16080.88