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

Equation of the Normal Line

$y - {e}^{5} = - \frac{1}{3 {e}^{5}} \left(x - 2\right)$

#### Explanation:

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 $\left({x}_{1} , {y}_{1}\right) = \left(2 , {e}^{5}\right)$

Solve for the slope $m$

$y ' = {e}^{{x}^{2} - x + 3} \cdot \left(2 x - 1\right)$

$m = {e}^{{2}^{2} - 2 + 3} \cdot \left(2 \cdot 2 - 1\right)$

$m = 3 \cdot {e}^{5}$

For the Normal Line

${m}_{n} = - \frac{1}{m} = - \frac{1}{3 \cdot {e}^{5}}$

Solve for the normal line

$y - {y}_{1} = {m}_{n} \left(x - {x}_{1}\right)$

$y - {e}^{5} = - \frac{1}{3 \cdot {e}^{5}} \left(x - 2\right)$

Kindly see the graph of $y - {e}^{5} = - \frac{1}{3 \cdot {e}^{5}} \left(x - 2\right)$ which is the red line and the blue curve is the $y = {e}^{{x}^{2} - x + 3}$

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