What is the equation of the normal line of $f(x)=x^3+6x^2-3x$ at $x=-1$ ?

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Answer 1 · 2018-03-29T07:51:40

$12y-x+14=0$

$f(x)=x^3+6x^2-3x$

to find the normal at $x=-1$

so the $y$ value

$y=f(-1)=(-1)^3+6(-1)^2-3(-1)$

$y=-1+6-3=2$

$:. (x_1,y_1)=(-1,2)---(1)$

we need the gradient at $x=-1$

the tangent gradient is

$f'(-1)$

and the normal gradient will be calculated from

$m_tm_n=-1$

$f(x)=x^3+6x^2-3x$

$f'(x)=3x^2+12x-3$

$m_t=f'(-1)=3(-1)^2+12(-1)-3$

$m_t=3-12-3=-12$

$=>m_n= -1/m_t=1/12 " from "(1)$

eqn of normal

$(y-y_1)=m(x-x_1)$

$(y- -1)=1/12(x-2)$

$12(y+1)=x-2$

$=>12y-x+14=0$

What is the equation of the normal line of f(x)=x^3+6x^2-3xf(x)=x^3+6x^2-3x at x=-1x=-1?