What is the equation of the normal line of f(x)=x2−xx−3 at x=4?

Explanation:First we evaluate the derivative of the function at x=4:f′(x)=(2x−1)(x−3)−(x2−x)(x−3)2 using the Quotient Rule.Evaluate it at x=4:f′(4)=((7⋅1)−12)1=−5=mThis is the slope of the TANGENT to our curve at x=4; to get the NORMAL we use m′=−1m=15 as slope.The point where our normal passes on the curve represented by our function, has coordinates:x0=4f(4)=y0=16−41=12and the equation of the line through this point with slope m' is given as:(y−y0)=m′(x−x0)y−12=15(x−4)y=15x+565

Expert answers to "What is the equation of the normal line..."

High School
Calculus and Analysis
Calculus 1
Derivatives

ds522stlk

Answered question

2022-04-08

What is the equation of the normal line of

f (x) = \frac{x^{2} - x}{x - 3}

at x=4?

Answer & Explanation

Austin Sherman

Beginner2022-04-09Added 12 answers

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

f^{'} (x) = \frac{(2 x - 1) (x - 3) - (x^{2} - x)}{{(x - 3)}^{2}}

using the Quotient Rule.
Evaluate it at x=4:

f^{'} (4) = \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^{'} = - \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 (4) = 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^{'} (x - x_{0})

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

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

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