What is the equation of the line that is normal to $f (x) = {(x + 4)}^{2} e^{x}$ $f(x)=(x+4)^2e^x$ at $x = - 3$ $x=-3$ ?

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Answer 1 · 2017-02-08T03:16:27

$\frac{1}{3} e^{3} x + y - e^{- 3} + e^{3} = 0$ $1/3e^3x+y-e^(-3)+e^3=0$ . See normal-inclusive Socratic graph

graph{(e^x(x+4)^2-y)(1/3e^3x+y-e^(-3)+e^3)=0 [-4, -2, -0.5, 0.5]}

y > 0.

As $x \to = \infty, y \to 0$ $xto=oo, y to 0$ , revealing that $y = 0 \leftarrow$ $y = 0 larr$ is the asymptote.

At $x = - 3, y = e^{- 3}$ $x =-3, y = e^(-3)$ .

So, the foot of the normal is $P (- 3, e^{- 3})$ $P(-3, e^(-3))$

$y'=e^x((x+4)^2+2(x+4))=3e^(-3)$ , at $x = -3$ .

The slope of the normal is $-1/(y')=-1/3e^3$

And so, the equation to the normal at P is

$y-e^(-3)=-1/3e^3(x+3)$ , giving

$1/3e^3x+y-e^(-3)+e^3=0$

What is the equation of the line that is normal to f(x)=(x+4)^2e^xf(x)=(x+4)2exf(x)=(x+4)^2e^x at x=-3 x=−3 x=-3 ?