Question

What is the equation of the line normal to `f(x)=(x-4)(x-2)+2x^2-3x+4` at `x=0`?

Answer 1

`y = 1/9x + 12`

Explanation:

Let's simplify the function first.

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

`color(white)(xxx) = x^2 - 6x + 8 + 2x^2 - 3x + 4`

`color(white)(xxx) = 3 x^2 - 9x + 12`

As next, to find the slope of the tangent and normal line, we need to compute the derivative of the function:

`f'(x) = 6x - 9`

Evaluating the derivative at `x=0` gives you the slope of the tangent at `x = 0`:

`m_t = f'(0) = -9`

The slope of the normal line at `x = 0` can be computed as

`m_n = - 1 / m_t = 1/9`

Now, we have the slope of the normal line and we have the `x` value of a point on this line: `x = 0`. Let's find the according `y` value by evaluating `f(0)`:

`f(0) = 3 * 0 - 9 * 0 + 12 = 12`

The normal line can be described via the line equation

`y = m_n * x + n`

Plug the values `m_n = 1/9`, `x = 0` and `y = 12` to find `n`:

`12 = 1/9 * 0 + n`

`=> n = 12`

So, the equation of the normal line is

`y = 1/9x + 12`