Question

What is the equation of the normal line of `f(x)=2x^3-x^2-3x+9` at `x=-1`?

Answer 1

Let's first find what point the tangent passes through, given `x = a`.

Explanation:

`f(x) = 2(-1)^3 - (-1)^2 - 3(-1) + 9`

`f(x) = -2 - 1 + 3 + 9`

`f(x) = 9`

Therefore, the tangent passes through `(-1, 9)`.

Now that we know this, we must differentiate the function.

By the power rule:

`f'(x) = 6x^2 - 2x - 3`

The slope of the tangent is given by evaluating `f(a)` inside the derivative, a being `x = a`.

`f'(-1) = 6(-1)^2 - 2(-1) - 3`

`f'(-1) = 6 + 2 - 3`

`f'(-1) = 5`

The slope of the tangent is `5`. The normal is always perpendicular to the tangent, so the slope will be the negative reciprocal of that of the tangent.

This means the slope of the normal line is `-1/5`. By point slope form, we can find the equation of the normal line:

`y - y_1 = m(x - x_1)`

`y - 9 = -1/5(x - (-1))`

`y - 9 = -1/5x - 1/5`

`y = -1/5x - 1/5 + 9`

`y = -1/5x + 44/5`

`:.` The equation of the normal line is `y = -1/5x + 44/5`.

Hopefully this helps!