Marshall Wolf

2022-04-07

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

convinto44re

Beginner2022-04-08Added 9 answers

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: $(y-{y}_{1})={m}_{\mathrm{norm}}(x-{x}_{1})$

Let's go through the process.

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

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

${m}_{\mathrm{norm}}=-\frac{1}{{m}_{\mathrm{tangent}}}=-\frac{1}{10}$

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

Substituting, we get the equation of the normal line.

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

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

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