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

Feb 15, 2016

$x - 6 y = - 52$

Explanation:

Given $f \left(x\right) = {\left(5 - x\right)}^{2}$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow f \left(x\right) = 25 - 10 x + {x}^{2}$
Its general slope is given by the derivative:
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 2 x - 10$

At $x = 2$ this slope will be
$\textcolor{w h i t e}{\text{XXX}} 2 \left(2\right) - 10 = - 6$
and a line normal to it (i.e. perpendicular to it) will have a slope
$\textcolor{w h i t e}{\text{XXX}} = - \left(\frac{1}{- 6}\right) = \frac{1}{6}$

At $x = 2 , f \left(2\right) = {\left(5 - 2\right)}^{2} = 9$
giving a point $\left(2 , 9\right)$

Using the slope-point form for the normal line
$\textcolor{w h i t e}{\text{XXX}} y - 9 = \frac{1}{6} \left(x - 2\right)$

We can simplify this as
$\textcolor{w h i t e}{\text{XXX}} 6 y - 54 = x - 2$
or (in standard form)
$\textcolor{w h i t e}{\text{XXX}} x - 6 y = - 52$
graph{(y-(5-x)^2)(x-6y+52)=0 [-7.83, 14.67, -0.585, 10.665]}