At what point on the given curve is the tangent line parallel to the line 5x - y = 5?
y = 4 + 2ex − 5x

Question

Emily Ammerman · Accepted Answer

#(\ln5, 14-5\ln5)#

Equation of given curve:

#y=4+2e^x-5x#

#\frac{d}{dx}y=\frac{d}{dx}(4+2e^x-5x)#

#\frac{dy}{dx}=2e^x-5#

The above derivative #\frac{dy}{dx}# shows the slope of tangent line at any point to the given curve.

It is given that the tangent to the curve is parallel to the line: #5x-y=5# or #y=5x-5# hence the slope of tangent will be equal to the slope of line i.e. #5#

hence, the slope of tangent #\frac{dy}{dx}# must be equal to #5# as follows

#\frac{dy}{dx}=5#

#2e^x-5=5#

#e^x=5#

#x=\ln5#

Substituting #x=\ln5# in equation of curve to get y-coordinate of point of contact as follows

#y=4+2e^{\ln5}-5\ln5#

#=4+2\cdot 5-5\ln 5#

#=14-5\ln5#

hence, the point of contact on the given curve is #(\ln5, 14-5\ln5)#

Isabella Ackerman · Answer

undefined To find the point on the given curve where the tangent line is parallel to the line 5x - y = 5, we need to find the derivative of the curve and set it equal to the slope of the given line.

The derivative of y = 4 + 2ex − 5x is dy/dx = 2ex - 5.

The slope of the line 5x - y = 5 can be determined by rearranging the equation to y = 5x - 5, which is in the form y = mx + b, where m represents the slope. Therefore, the slope of the line is 5.

Setting the derivative equal to the slope of the line, we have 2ex - 5 = 5.

Solving this equation for x, we get x = ln(5/2).

To find the corresponding y-coordinate, we substitute this value of x back into the original equation y = 4 + 2ex − 5x.

Therefore, the point on the given curve where the tangent line is parallel to the line 5x - y = 5 is (ln(5/2), 4 + 2e^(ln(5/2)) − 5(ln(5/2))).

At what point on the given curve is the tangent line parallel to the line 5x - y = 5? y = 4 + 2ex − 5x