Question

What is the equation of the normal line of `f(x)=-x^4+2x^3+2x^2-x-4` at `x=1`?

Answer 1

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

Explanation:

To build the equation of the normal line, you need a point on the normal line and its slope.

1) Point on the normal line

As you are searching for the normal line at `x = 1`, your `x` coordinate is `x = 1`.

To find the `y` value of the point, you need to evaluate `f(x)` for `x = 1`:

`f(1) = - 1^4 + 2 * 1^3 + 2 * 1^2 - 1 - 4 = - 1 + 2 + 2 - 1 - 4 = -2`

Thus, the point is `P (1 | -2)`

2) Slope of the normal line

To compute the slope of the normal line, first of all, you need the derivative of `f(x)`:

`f'(x) = - 4 x^3 + 6 x^2 + 4 x - 1`

Now, you need to evaluate the derivative at `x = 1`. This will give you the slope of the tangent at `x = 1`.

`f'(1) = - 4 * 1^3 + 6 * 1^2 + 4 * 1 - 1 = - 4 + 6 + 4 - 1 = 5`

Thus, the slope of the tangent at `x = 1` is `m_t = 5`.

We know that the slope of the normal line `m_n` can be computed as

`m_n = - 1 / m_t = - 1 /5`

3) Computing the equation of the normal line

Now that we have the point and the slope, we can build the equation.

Any line has the form

`y = m * x + n`

We already have `y = -2`, `m = -1/5` and `x = 1`. Let's find `n`:

`-2 = - 1 /5 * 1 + n`

`-2 = - 1/5 + n`

... add `1/5` on both sides...

`-9 /5 = n`

Thus, the equation of the normal line is

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