Question

What is the equation of the normal line of `f(x)=x^3e^-x+x^3` at `x=-5`?

Answer 1

`y_(t)=color(red)m_tx+color(red)b_t`
The normal equation is `y_(n)= color(blue)m_nx+color(blue)b_n`
a) Where `color(red)m_t = -1/color(blue)m_n`
b) `color(red)b_t = color(blue)b_n`

Explanation:

the slope of the normal line is the negative reciprocal of the slope of the tangent line. so what you need to don is :
1) find the the slope of f(x) at `x = -5 => f'(-5)`
a) `f'(x)=3x^2e^-x - x^3e^-x + 3x^2`
b) `f'(-5)=y=75e^5 + 125e^5-125`
2) find the equation of line, equation of tangent at (-5, f(5))
a) Tangent equation, `y_(t)=color(red)m_tx+color(red)b_t`
b) `color(red)m_t = f'(-5)`
c) `y_(t)-f(-5)=color(red)f'(-5)(x+5)+color(red)b_t` solve for b
4) The normal equation is `y_(n)= color(blue)m_nx+color(blue)b_n`
a) Where `color(red)m_t = -1/color(blue)m_n`
b) `color(red)b_t = color(blue)b_n`

There is some algebra that you need to, which I leave to you. the recipe for solving is there for you for now and future problems...

Good luck