# What is the equation of the normal line of f(x)=-x^4+4x^3-x^2+5x-6 at x=2?

May 29, 2016

Slope of normal is $x + 17 y - 818 = 0$

#### Explanation:

As the function is $f \left(x\right) = - {x}^{4} + 4 {x}^{3} - {x}^{2} + 5 x - 6$, the slope of tangent at any point will be the value of $f ' \left(x\right)$ at that point.

As $f ' \left(x\right) = - 4 {x}^{3} + 12 {x}^{2} - 2 x + 5$, the slope of the tangent at $x = 2$ will be

$- 4 \cdot {2}^{3} + 12 \cdot {2}^{2} - 2 \cdot 2 + 5 = - 32 + 48 - 4 + 5 = 17$.

And slope of normal would be $- 1 \div 17 = - \frac{1}{17}$

Note that value of function at $x = 2$ is $f \left(x\right) = {2}^{4} + 4 \cdot {2}^{3} - {2}^{2} + 5 \cdot 2 - 6$

= $16 + 32 - 4 + 10 - 6 = 48$

Hence, slope of normal is $- \frac{1}{17}$ and it passes through $\left(2 , 48\right)$

its equation of normal is $y - 48 = - \frac{1}{17} \left(x - 2\right)$ or $17 \left(y - 48\right) = - x + 2$ or

$x + 17 y - 818 = 0$