Question

Find the equation to the tangent line to the curve at the specified point:-F(x) = - 2/3(x^3) + x^2 + 6x – 4at the point: (3, 5)

Answer 1

Given:

`\begin{gathered} f(x)=-(2)/(3)x^3+x^2+6x-4 \\ \text{ Point = (3,5)} \end{gathered}`

The general equation of tangent line is:

`y-y_1=m(x-x_1)`

Where,

`\begin{gathered} (x_1,y_1)=\text{ Point} \\ m=\text{ Slope} \end{gathered}`

Solpe of line is:

`\begin{gathered} m=f^(\prime)(x) \\ \end{gathered}`
`\begin{gathered} f(x)=-(2)/(3)x^3+x^2+6x-4 \\ f^(\prime)(x)=-(2)/(3)(3x^2)+2x+6 \\ x=3;y=5 \\ f^(\prime)(x)=-2x^2+2x+6 \\ f^(\prime)(3)=-2(3)^2+2(3)+6 \\ f^(\prime)(3)=-18+6+6 \\ f^(\prime)(3)=-6 \end{gathered}`

slope of tangent line is -6

then equation of tengent line is:

`\begin{gathered} y-y_1=m(x-x_1) \\ m=-6 \\ (x_1,y_1)=(3,5) \\ so, \\ y-5=-6(x-3) \\ y-5=-6x+18 \\ y+6x=18+5 \\ y+6x=23 \end{gathered}`