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

1 Answer
Feb 12, 2016

#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