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

Oct 28, 2016

$y = - 10 x - 25$

#### Explanation:

The equation of a straight line is determined by computing its $\textcolor{red}{s l o p e}$ and the point it passes through name it $\left(\textcolor{p u r p \le}{{x}_{1} , {y}_{1}}\right)$

This line is normal to $f \left(x\right)$ at $\textcolor{p u r p \le}{x = - 5}$ so it passes through the point$\left(\textcolor{p u r p \le}{- 5 , f \left(- 5\right)}\right)$

$\textcolor{p u r p \le}{f \left(- 5\right)} = {\left(- 5\right)}^{2} = 25$

This line passes through the point $\left(\textcolor{p u r p \le}{- 5 , 25}\right)$

The $\textcolor{red}{s l o p e}$ at $x = - 5$ is determined by computing $\textcolor{red}{f ' \left(- 5\right)}$
$f \left(x\right)$ is differentiated by using power rule differentiation
$\left({x}^{n}\right) ' = n \left({x}^{n - 1}\right)$

$f ' \left(x\right) = 2 x$
$\textcolor{red}{f ' \left(- 5\right) = 2 \left(- 5\right) = - 10}$

The equation is:

$y - {y}_{1} = \textcolor{red}{s l o p e} \left(x - {x}_{1}\right)$

$y - f \left(- 5\right) = - 10 \left(x - \left(- 5\right)\right)$

$y - 25 = - 10 \left(x + 5\right)$

$y = - 10 x - 50 + 25$

$y = - 10 x - 25$