What is the equation of the line normal to  f(x)=(x+5)(x+2)+2x^2-8x+2 at  x=0?

Jun 1, 2016

It is $y = x + 12$.

Explanation:

The strategy is to find first the slope of the tangent line than we can easily find the orthogonal.

The slope of the tangent is given by the derivative evaluate in $x = 0$
Before doing the derivative I do the product between the parenthesis and reorder the terms

$f \left(x\right) = \left(x + 5\right) \left(x + 2\right) + 2 {x}^{2} - 8 x + 2 = {x}^{2} + 2 x + 5 x + 10 + 2 {x}^{2} - 8 x + 2$
$= 3 {x}^{2} - x + 12$.

Now for the derivative

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} 3 {x}^{2} - x + 12 = 6 x - 1$
then when $x = 0$ we have
$f ' \left(0\right) = - 1$.

This is the the slope of the tangent line passing from $x = 0$.

The slope of the normal line is the opposite inverse of this:

$m = - \frac{1}{-} 1 = 1$.
We need now to find the intercept and to do this we impose the passage for the point $\left(0 , f \left(0\right)\right)$.
$f \left(0\right) = 3 \cdot {0}^{2} - 0 + 12 = 12$, so the normal line passes from $\left(0 , 12\right)$.

The equation is
$y = m x + q$ with $m = 1$ it is
$y = x + q$ and not I substitute the point $\left(0 , 12\right)$
$12 = q$

Then the final equation of the normal line is

$y = x + 12$.