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2022-04-08

What is the equation of the normal line of $f\left(x\right)=\frac{{x}^{2}-x}{x-3}$ at x=4?

Austin Sherman

Beginner2022-04-09Added 12 answers

Explanation:

First we evaluate the derivative of the function at x=4:

$f}^{\prime}\left(x\right)=\frac{(2x-1)(x-3)-({x}^{2}-x)}{{(x-3)}^{2}$ using the Quotient Rule.

Evaluate it at x=4:

${f}^{\prime}\left(4\right)=\frac{((7\cdot 1)-12)}{1}=-5=m$

This is the slope of the TANGENT to our curve at x=4; to get the NORMAL we use$m}^{\prime}=-\frac{1}{m}=\frac{1}{5$ as slope.

The point where our normal passes on the curve represented by our function, has coordinates:

${x}_{0}=4$

$f\left(4\right)={y}_{0}=\frac{16-4}{1}=12$

and the equation of the line through this point with slope m' is given as:

$(y-{y}_{0})={m}^{\prime}(x-{x}_{0})$

$y-12=\frac{1}{5}(x-4)$

$y=\frac{1}{5}x+\frac{56}{5}$

First we evaluate the derivative of the function at x=4:

Evaluate it at x=4:

This is the slope of the TANGENT to our curve at x=4; to get the NORMAL we use

The point where our normal passes on the curve represented by our function, has coordinates:

and the equation of the line through this point with slope m' is given as:

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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