Malachi Mullins

2022-04-10

What is the equation of the line that is normal to $f\left(x\right)={\left(x-3\right)}^{2}+{x}^{2}-6x$ at x=0?

carlosegundoacyg

Explanation:
For given f(x), f '(x)= 2(x-3) +2x-6. Slope of tangent to f(x), at x=0 would be = -12. Therefore slope of the normal line would be $\frac{1}{12}$. At x=0, f(x)= 9. Hence equation of normal line at this point would be $y-9=\frac{1}{12}x$ (slope-intercept from of line).
Result:
$y-9=\frac{1}{12}x$

WigwrannyErarmbmk

Given:
$f\left(x\right)={x}^{2}-6x+{\left(x-3\right)}^{2}$ and ${x}_{0}=0$.
Find the value of the function at the given point: ${y}_{0}=f\left(0\right)=9$.
The slope of the normal line at $x={x}_{0}$ is the negative reciprocal of the derivative of the function,
evaluated at $x={x}_{0}:M\left({x}_{0}\right)=-\frac{1}{{f}^{\prime }\left({x}_{0}\right)}$
Find the derivative: ${f}^{\prime }\left(x\right)={\left({x}^{2}-6x+{\left(x-3\right)}^{2}\right)}^{\prime }=4\left(x-3\right)$
Hence, $M\left({x}_{0}\right)=-\frac{1}{{f}^{\prime }\left({x}_{0}\right)}=-\frac{1}{4\left({x}_{0}-3\right)}$
Next, find the slope at the given point.
$m=M\left(0\right)=\frac{1}{12}$
Finally, the equation of the normal line is $y-{y}_{0}=m\left(x-{x}_{0}\right)$
Plugging the found values, we get that $y-9=\frac{x-0}{12}$
Or, more simply: $y=\frac{x}{12}+9$

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