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

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Answer 1 · 2016-09-01T20:23:12

$x + 13 y + 248 = 0$ $x+13y+248=0$ .

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

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

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

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

We have, $f'(x)=3x^2-2x+8 rArr f'(-1)=3+2+8=13$ .

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

$C$ at the pt. $A$ .

$:." the slope of tgt. at "A" is "13$ .

Normal is $bot$ to tgt., so, its slope is $-1/13$ , it passes thro. $A$ .

Hence, eqn. of normal is $: y+19=-1/13(x+1)$ , or,

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

Enjoy Maths.!

What is the equation of the normal line of f(x)=x^3-x^2+8x-9f(x)=x3−x2+8x−9f(x)=x^3-x^2+8x-9 at x=-1x=−1x=-1?