Question

What is the equation of the normal line of `f(x)=-2x^3-8x^2-2x` at `x=-1`?

Answer 1

`y = -1/8 x - 33/8`

Explanation:

To compute the equation of the normal line, you need:

• the slope of the normal line
• one point that is on the normal line

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1) Let's start with the point.

You already know that the normal line intercepts your function `f(x)` at `x = -1`.

So, the only thing left to do is computing the `y` value for `x = -1`:

`f(-1) = -2 * (-1) - 8 * 1 - 2 * (-1) = 2 - 8 + 2 = -4`

So, your point is `(-1 | -4)`.

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2) Now, let's take care of the slope . To do so, you need to differentiate `f(x)` and then compute the value of `f'(-1)`.

`f'(x) = -6 x^2 - 16 x - 2`

`f'(-1) = -6 * 1 - 16 * (-1) - 2 = -6 + 16 - 2 = 8`

This means that the slope of the tangent line at `x = -1` is `m_t = 8`.
The slope of the normal line is `m_n = - 1 / m_t = - 1/8`.

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3) We have the point and the slope. Let's compute the equation of the normal line.

The equation of any line is `y = mx + n`.

We know that `m = -1/8` and for `x` and `y` we can plug the `x` and `y` values of the point that we know is on the line: `x = -1` and `y = -4`.

With this knowledge, we can compute `n`:

`-4 = -1/8 * (-1) + n`

`<=> -4 = 1/8 + n`

`<=> n = -33/8`

So, the equation of the normal line is

`y = -1/8 x - 33/8`

Hope that this helped!