Find the equation of the line tangent to the curve

Question

F(x) = 3-2x-4x^2 at x=2 

Answer 1

 

In order to get the slope of the line tangent to the curve,

we take the first order derivative of F(x) which is given as follows:

 

     F'(x) = (3 - 2x - 4x²)'

             = -2 - 8x

 

And, the slope of the line at x = 2 can be calculated as follows:

 

     m = F'(2)

         = -2 - 8·2

         = -18

 

From the equation of the original curve,

the y-coordinate of the point the line passes through can be calculated as follows:

 

     F(2) = 3 - 2·2 - 4·2²

            = -17

 

The line passes through the point (2, -17) with the slope m of -18.

 

So, the equation of the line is given as follows:

 

     y - (-17) = -18 (x - 2)

 

The final answer, the equation of the line, is

 

     y = -18x + 19          

 

 

 

 

 

Answer 2

First we need to find F(2) so that we have a y-value for our equation

 

F(2) = 3-2(2)-4(2)^2 = -19

 

then we need to take the first derivative.

 

F'(x) = 2 - 8x

 

then find F'(2) which will be the slope of the tangent line.

 

F'(2) = 2 - 8(2) = -14

 

so our slope is m=-14 and our point is at (2,-19)

 

so if we put this into the point slope form (y-y1)=m(x-x1) we get

 

(y+19)=-14(x-2)