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

May 31, 2016

$78 y = - x + 4680$

#### Explanation:

First find the slope of the tangent at the given point by differentiating and then substituting in $x = 4$

$f ' \left(x\right) = 8 {\left(x - 2\right)}^{3} + 3 {\left(x - 1\right)}^{2} + 2 \left(x - 3\right)$

At $x = 4$ this is $m = 8 \cdot {2}^{3} + 3 \cdot {2}^{2} + 2 \cdot 1 = 64 + 12 + 2 = 78$

The slope of the normal line is $- \frac{1}{m} = - \frac{1}{78}$

The equation of the normal is then $y = - \frac{1}{78} x + c$

To calculate $c$, find the value of the original function at the given point and then substitute into the equation of the normal.

$f \left(4\right) = 2 \cdot {2}^{4} + {3}^{3} + {1}^{2} = 32 + 27 + 1 = 60$

$\therefore 60 = - \frac{1}{78} \cdot 4 + c$

$c = 60 + \frac{4}{78} = \frac{4680}{78}$

$y = - \frac{1}{78} x + \frac{4680}{78}$

$\therefore 78 y = - x + 4680$