# What is the equation of the normal line of f(x)=x^3-x^2+8x-9 at x=-1?

Sep 1, 2016

$x + 13 y + 248 = 0$.

#### Explanation:

$f \left(x\right) = {x}^{3} - {x}^{2} + 8 x - 9$

rArr f(-1)=-1-1-8-9=-19.

Therefore, we require the eqn. of normal to the curve

$C : y = f \left(x\right) = {x}^{3} - {x}^{2} + 8 x - 9$ at the pt.$A \left(- 1 , - 19\right)$.

We have, $f ' \left(x\right) = 3 {x}^{2} - 2 x + 8 \Rightarrow f ' \left(- 1\right) = 3 + 2 + 8 = 13$.

But, we know that, f"(-1)# gives the slope of tgt. line to the Curve

$C$ at the pt.$A$.

$\therefore \text{ the slope of tgt. at "A" is } 13$.

Normal is $\bot$ to tgt., so, its slope is $- \frac{1}{13}$, it passes thro. $A$.

Hence, eqn. of normal is $: y + 19 = - \frac{1}{13} \left(x + 1\right)$, or,

$13 y + 247 + x + 1 = 0 , i . e . , x + 13 y + 248 = 0$.

Enjoy Maths.!