What is the equation of the normal line of #f(x)= e^(x^2-x+3)# at #x = 2#?

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Answer 1 · 2016-05-05T04:57:38

Equation of the Normal Line

#y-e^5=-1/(3e^5)(x-2)#

Solve for the ordinate first

#y=e^(x^2-x+3)# at #x=2#

#y=e^(2^2-2+3)#

#y=e^5#

We now have #(x_1, y_1)=(2, e^5)#

Solve for the slope #m#

#y'=e^(x^2-x+3)*(2x-1)#

#m=e^(2^2-2+3)*(2*2-1)#

#m=3*e^5#

For the Normal Line

#m_n=-1/m=-1/(3*e^5)#

Solve for the normal line

#y-y_1=m_n(x-x_1)#

#y-e^5=-1/(3*e^5)(x-2)#

Kindly see the graph of #y-e^5=-1/(3*e^5)(x-2)# which is the red line and the blue curve is the #y=e^(x^2-x+3)#

desmos

God bless....I hope the explanation is useful.