# What is the equation of the line that is normal to f(x)=4x^3-3x^2+5x-2 at  x=1 ?

Jan 26, 2016

$11 y = - x + 44$

#### Explanation:

The slope of the normal at a given point is the negative inverse of the slope of the function at that point. i.e. if the slope of the function is $m$ then the slope of the normal is $- \frac{1}{m}$

To find the slope, differentiate the function.

$f ' \left(x\right) = 12 {x}^{2} - 6 x + 5$

At $x = 1$ this equals $12 - 6 + 5 = 11$

Therefore the slope of the normal is $- \frac{1}{11}$

The calculate the constant $c$ in the standard lien equation $y = m x + c$, substitute $x = 1$ back into the original function $f \left(x\right)$
$= 4 \cdot {1}^{3} - 3 \cdot {1}^{2} + 5 \cdot 1 - 2 = 4$

Therefore the normal is $y = \left(- \frac{1}{11}\right) x + 4$
$11 y = - x + 44$