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

Question

Ezekiel Alexander · Accepted Answer

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

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)#

Christopher Allers · Answer

undefined To find the tangent line parallel to the line 2x+y=7 for the curve y=e^(-4x), we need to determine the slope of the tangent line. The slope of the given line is -2.

To find the slope of the tangent line for the curve, we take the derivative of the function y=e^(-4x) with respect to x. The derivative of e^(-4x) is -4e^(-4x).

Setting -4e^(-4x) equal to -2 (the slope of the given line), we can solve for x.

-4e^(-4x) = -2

Dividing both sides by -4, we get:

e^(-4x) = 1/2

Taking the natural logarithm of both sides, we have:

-4x = ln(1/2)

Simplifying, we find:

x = -ln(1/2)/4

Now that we have the x-coordinate, we can substitute it back into the original function y=e^(-4x) to find the corresponding y-coordinate.

y = e^(-4(-ln(1/2)/4))

Simplifying further, we get:

y = e^(ln(1/2))

y = 1/2

Therefore, the point of tangency is (-ln(1/2)/4, 1/2).

Using the point-slope form of a line, we can write the equation of the tangent line parallel to 2x+y=7 as:

y - 1/2 = -2(x + ln(1/2)/4)