What is the equation of the normal line of #f(x)=x^3*(3x - 1) # at #x=-2 #?

Answer 1

#y=1/108x-3135/56#

The normal line to a tangent is perpendicular to the tangent. We can find the slope of the tangent line using the derivative of the original function, then take its opposite reciprocal to find the slope of the normal line at the same point.

#f(x)=3x^4-x^3#
#f'(x)=12x^3-3x^2#
#f'(-2)=12(-2)^3-3(-2)^2=12(-8)-3(4)=-108#
If #-108# is the slope of the tangent line, the slope of the normal line is #1/108#.
The point on #f(x)# that the normal line will intersect is #(-2,-56)#.

We can write the equation of the normal line in point-slope form:

#y+56=1/108(x+2)#

In slope-intercept form:

#y=1/108x-3135/56#
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Answer 2

The equation of the normal line of f(x)=x^3*(3x - 1) at x=-2 is y = -4x - 10.

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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.

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.

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.

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.

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