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What is the equation of the normal line of #f(x)=12x^3-4x^2-5x# at #x=-2#?

Answer 1

#y = -1/155x - (15812)/155#

Normal line will be perpendicular to the tangent line. As we know that the product of perpendicular gradients is always #-1# we can find the gradient of the normal from the gradient of the tangent.

We compute the slope of the tangent by evaluating the first derivative:

#f'(x) = 36x^2 - 8x - 5#

#f'(-2) = 144 + 16 - 5 = 155#

So the slope of the normal (#m_n#) can be found by:

#m_n*155 = -1 implies m_n = -1/155#

To calculate the equation of the normal line we use

#y-b = m(x-a)#

To use this, we need a point on the line. We know that x = -2 is on the line, so we evaluate the original function at this point to get :

#f(-2) = -102# hence:

#y - (-102) = -1/155(x - (-2))#

#y+102 = -1/155x -2/155#

#y = -1/155x - (15812)/155#

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Answer 2

Samantha Adkison

The equation of the normal line of f(x)=12x^3-4x^2-5x at x=-2 is y = -23x - 26.

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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.