For the curve #y=e^(-4x)#, how do you find the tangent line parallel to the line #2x+y=7#?

1 Answer
Jul 28, 2015

I found: #y=-2x-1/2[ln(1/2)-1]#

Explanation:

The tangent line will have slope equal to the derivative of your function:
#y'=-4e^(-4x)#
but it has to be parallel to the line #2x+y=7# or #y=-2x+7# which has slope #=-2#;
So the two slopes must be equal:
#-4e^(-4x)=-2#
#e^(-4x)=1/2#
#-4x=ln(1/2)#
#x=-1/4ln(1/2)#
#x=0.173#
so this is the #x# coordinate of the tangence point the other being:
#y=e^(-4ln(1/2)/-4)=1/2#
So your tangent line has slope #m=-2# and passes therough:
#x_0=-1/4ln(1/2)#
#y_0=1/2#
The equation of this line will be:
#y-y_0=m(x-x_0)#
#y-1/2=-2[x+1/4ln(1/2)]#
#y=-2x-1/2ln(1/2)+1/2#
#y=-2x-1/2[ln(1/2)-1]#

Graphically:
with:
#y=e^(-4x)#
#y1=-2x+7# (the given line) and:
#y2=-2x-1/2(ln(1/2)-1)#
enter image source here