What is the equation of the normal line of f(x)=x2 at x=5?

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convinto44re · Accepted Answer

The normal line is the line perpendicular to the tangent line. Then, the steps are as follows:
Find the slope $m_{tangent}$ of tangent line
-Find the first derivative of the function
-Evaluate the first derivative at the desired point
Find the slope $m_{norm}$ of the normal line
-Let $m_{norm} = - \frac{1}{m_{tangent}}$
Find the line with slope $m_{norm}$ passing through the given point
-Using point-slope form: $(y - y_{1}) = m_{norm} (x - x_{1})$
Let's go through the process.
$f^{'} (x) = \frac{d}{d x} x^{2} = 2 x$
$m_{tangent} = f^{'} (5) = 2 (5) = 10$
$m_{norm} = - \frac{1}{m_{tangent}} = - \frac{1}{10}$
$(x_{1}, y_{1}) = (5, f (5)) = (5, 25)$
Substituting, we get the equation of the normal line.
$y - 25 = - \frac{1}{10} (x - 5)$
$\Rightarrow y = - \frac{1}{10} x + \frac{51}{2}$

What is the equation of the normal line

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