# What is the equation of the normal line of f(x)=x^2  at x=5 ?

Dec 4, 2015

$y = - \frac{1}{10} x + \frac{51}{2}$

#### Explanation:

The normal line is the line perpendicular to the tangent line. Then, the steps are as follows:

• Find the slope ${m}_{\text{tangent}}$ of tangent line
-Find the first derivative of the function
-Evaluate the first derivative at the desired point

• Find the slope ${m}_{\text{norm}}$ of the normal line
-Let ${m}_{\text{norm" = -1/m_"tangent}}$

• Find the line with slope ${m}_{\text{norm}}$ passing through the given point
-Using point-slope form: $\left(y - {y}_{1}\right) = {m}_{\text{norm}} \left(x - {x}_{1}\right)$

Let's go through the process.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {x}^{2} = 2 x$

${m}_{\text{tangent}} = f ' \left(5\right) = 2 \left(5\right) = 10$

${m}_{\text{norm" = -1/m_"tangent}} = - \frac{1}{10}$

$\left({x}_{1} , {y}_{1}\right) = \left(5 , f \left(5\right)\right) = \left(5 , 25\right)$

Substituting, we get the equation of the normal line.

$y - 25 = - \frac{1}{10} \left(x - 5\right)$

$\implies y = - \frac{1}{10} x + \frac{51}{2}$