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

Answer 1

Emilia Aguilar

The equation of the normal line is #y=-1/14x-85/12#.

We have the function, #f(x)=x^3-x^2#.

At #x=-2#, #f(x)=(-2)^3-(-2)^2#

#=-8-4#

#=-12#

So, the point is at #(-2,-12)#.

First, we find the slope of the tangent line at #x=-2#, and that's the derivative of the function.

#f'(x)=3x^2-2x#

So at #x=-2#,

#f'(-2)=3*(-2)^2-2*-2#

#=3*4+4#

#=14#

Now, the normal line is perpendicular to the tangent line, and we know that the slope of the normal line is the negative reciprocal of the tangent line, as

#m_1m_2=-1#, where #m_1,m_2# are the gradients of the tangent line and normal line, respectively.

And so, the slope becomes, #-1/14#.

The point-slope form states that,

#y-y_0=m(x-x_0)#

And so,

#y-(-12)=-1/14(x-(-2))#

#y+12=-1/14(x+2)#

#y+12=-1/14x-1/7#

#y=-1/14x-1/7-12#

#=-1/14x-85/12#

Here is a graph of the line:

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Answer 2

Aurora Alvarado

The equation of the line normal to f(x)=x^3-x^2 at x=-2 is y = -12x - 16.

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