Find the equation of the tangent to the curve y = 3x - x^3 at x = 2 .

To determine the equation of the tangent line to the graph
of y, we'll have to determine the derivative of y at x =
2.

f'(2) = lim [f(x) -
f(2)]/(x-2)

f(2) = 3*2 -
2^3

f(2) = 6 - 8

f(2) =
-2

lim [f(x) - f(2)]/(x-2) = lim (3x - x^3 +
2)/(x-2)

We'll substitute x by
2:

lim (3x - x^3 + 2)/(x-2) = (6-8+2)/(2-2) =
0/0

Since we've obtained in indeterminacy acse, we'll apply
L'Hospital rule:

lim (3x - x^3 + 2)/(x-2) = lim (3x - x^3 +
2)'/(x-2)'

lim (3x - x^3 + 2)'/(x-2)' = lim
(3-3x^2)/1

We'll substitute x by
2:

lim (3-3x^2)/1 = 3 - 3*4 = 3-12 =
-9

f'(2) = -9

But the
derivative of y at x = 2 is the slope of the tangent line to the curve
y.

m = -9

Now, we'll write the
equation of the tangent line, whose slope is m=-9 and it passes through the point that
has x coordinate = 2.We'll compute the y coordinate of this
point:

f(2) = 3x - x^3

f(2) =
6 - 8

f(2) = -2

The equation
of the tangent line is:

y - (-2) = m(x -
2)

y + 2 = -9(x - 2)

We'll
remove the brackets:

y + 2 = -9x +
18

y + 9x + 2 - 18 =
0

y + 9x - 16 =
0