What is the equation of the line tangent to # f(x)=(x-1)^2(2x+4)# at # x=0 #?

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Answer 1 · 2017-12-04T01:03:48

#y=-6x+4#

Solve for #y# by plugging #x=0# into the given equation.

#f(0)=(-1)^2(4)=4#

Find the slope of the tangent line by solving for the second derivative. Use the product rule.

#f'(x)=2(x-1)^2+(2x+4)2(x-1)#
#f'(x)=2(x-1)^2+(2x+4)(2x-2)#

Plug in #x=0# to find the slope of the tangent line at (0, 4) point.

#f'(0)=2(-1)^2+(4)(-2)#
#f'(0)=2-8=-6#

With your knowledge of the slope and (x,y) coordinate, use the following formula to find the equation of the tangent line.

#y-y_1=m(x-x_1)#
#y-4=-6(x-0)#
#y=-6x+4#