Question

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`

Explanation:

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`