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

Dec 3, 2015

$y = \frac{1}{9} x + 12$

#### Explanation:

Let's simplify the function first.

$f \left(x\right) = \left(x - 4\right) \left(x - 2\right) + 2 {x}^{2} - 3 x + 4$

$\textcolor{w h i t e}{\times x} = {x}^{2} - 6 x + 8 + 2 {x}^{2} - 3 x + 4$

$\textcolor{w h i t e}{\times x} = 3 {x}^{2} - 9 x + 12$

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

$f ' \left(x\right) = 6 x - 9$

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

${m}_{t} = f ' \left(0\right) = - 9$

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

${m}_{n} = - \frac{1}{m} _ t = \frac{1}{9}$

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 $f \left(0\right)$:

$f \left(0\right) = 3 \cdot 0 - 9 \cdot 0 + 12 = 12$

The normal line can be described via the line equation

$y = {m}_{n} \cdot x + n$

Plug the values ${m}_{n} = \frac{1}{9}$, $x = 0$ and $y = 12$ to find $n$:

$12 = \frac{1}{9} \cdot 0 + n$

$\implies n = 12$

So, the equation of the normal line is

$y = \frac{1}{9} x + 12$