Please help me solve this calculus questions.

Question

Determine the equation of the tangent to the curve defined by y = 2e^3x that is parallel to the line defined by - 6x + y = 9

Answer 1

Given:

f(x) = 2e^3x

Tangent Line is Parallel to: -6x + y = 9

Find:

Tangent Line to f = T(x) = ?

Solution:

f'(x) = 6e3x

d/dx[ -6x + y ] = d/dx[ 9 ]

-6 + y' = 0

y' = 6

To be PARALLEL, the slopes must be equal

6 = 6 e^3x

ln( e^3x ) = ln( 1 )

3x = 0

∴ x = 0

f( 0 ) = 2 exp( 3( 0 ) )

∴ f( 0 ) = 2

y - 2 = 6( x - 0 )

∴ T(x) = 6x - 2

I hope this helps! Message me in the comments with any questions, comments, or concerns!