Marshall Wolf

2022-04-07

What is the equation of the normal line of $f\left(x\right)={x}^{2}$ at x=5?

convinto44re

The normal line is the line perpendicular to the tangent line. Then, the steps are as follows:
Find the slope ${m}_{\mathrm{tangent}}$ of tangent line
-Find the first derivative of the function
-Evaluate the first derivative at the desired point
Find the slope ${m}_{\mathrm{norm}}$ of the normal line
-Let ${m}_{\mathrm{norm}}=-\frac{1}{{m}_{\mathrm{tangent}}}$
Find the line with slope ${m}_{\mathrm{norm}}$ passing through the given point
-Using point-slope form: $\left(y-{y}_{1}\right)={m}_{\mathrm{norm}}\left(x-{x}_{1}\right)$
Let's go through the process.

${m}_{\mathrm{tangent}}={f}^{\text{'}}\left(5\right)=2\left(5\right)=10$
${m}_{\mathrm{norm}}=-\frac{1}{{m}_{\mathrm{tangent}}}=-\frac{1}{10}$
$\left({x}_{1},{y}_{1}\right)=\left(5,f\left(5\right)\right)=\left(5,25\right)$
Substituting, we get the equation of the normal line.
$y-25=-\frac{1}{10}\left(x-5\right)$
$⇒y=-\frac{1}{10}x+\frac{51}{2}$

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