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

Answer 1

Emily Ammerman

#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#

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Answer 2

Elijah Abram

The equation of the line tangent to f(x)=(x-1)^2(2x+4) at x=0 is y = 4x + 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.