# What is the equation of the line normal to  f(x)=-(x+6)(x+3)+4x^2-8x+2 at  x=0?

Mar 27, 2017

$y = - \frac{1}{17} x - 16$

#### Explanation:

To find the slope of the normal line, we need to differentiate to get $f ' \left(x\right)$ , then plug in zero to find slope of tangent line at $x = 0$, then finally find the opposite reciprocal .

$f \left(x\right) = - \left(x + 6\right) \left(x + 3\right) + 4 {x}^{2} - 8 x + 2$

Differentiate using the product rule and power rule:
$f ' \left(x\right) = - \left[\left(x + 6\right) \left(1\right) + \left(1\right) \left(x + 3\right)\right] + 8 x - 8$

$f ' \left(x\right) = - x - 6 - x - 3 + 8 x - 8$

$f ' \left(x\right) = 6 x - 17$

Plug in zero:
$f ' \left(0\right) = 6 \left(0\right) + 17 = 17$

Find opposite reciprocal of tangent line slope to get normal line slope:
$\text{Slope of normal line} = - \frac{1}{17}$

Equation of normal line (point-slope form):
$f \left(0\right) = - 16$
$y + 16 = - \frac{1}{17} \left(x - 0\right)$

$y = - \frac{1}{17} x - 16$