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.

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