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

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Answer 1 · 2017-01-25T11:24:01

$x - 2 y + 7 = 0$ $x-2y+7=0$ . See the normal-inclusive Socratic graph.

$f (- 1) = - 2 + 3 + 2 = 3$ $f(-1)=-2+3+2=3$ . So, the foot of the normal P is $(- 1, 3)$ $(-1, 3)$ .

#f'=6x^2+6x-2=6-6-2=-2.

Slope of the normal #= -1/f'=1/2, at P(-1, 3).

So, the equation to the normal at P is

$y - 3 = \frac{1}{2} (x + 1)$ $y-3=1/2(x+1)$ , giving

$x - 2 y + 7 = 0$ $x-2y+7=0$

graph{(2x^3+3x^2-2x-y)(x-2y+7)((x+1)^2+(y-3)^2-.01)=0 [-10, 10, -5, 5]}

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