Find out the answer to "What is the equation of the normal line..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Rachael Velasquez

Answered question

2022-04-10

What is the equation of the normal line of

f (x) = 5 x^{4} + 3 x^{3} + x^{2} - 2 x - 4

at x=1?

Answer & Explanation

Emily Green

Beginner2022-04-11Added 14 answers

Differentiate the function to find its gradient.

f^{'} (x) = 20 x^{3} + 9 x^{2} + 2 x - 2

Use the point x=1 to find the gradient of the curve at that point.
f'(1)=20+9+2-2=29
The gradient of the normal is the negative reciprocal of the gradient of the tangent at a point.
So the gradient of the normal at x=1 is

m = - \frac{1}{29}

To find the coordinates of the point the normal passes through, substitute x=1
into f(x)
f(1)=5+3+1-2-4=3
So coordinates are

(x_{1}, y_{1}) \Rightarrow (1, 3)

Equation of a line is:

y - y_{1} = m (x - x_{1})

y - 3 = - \frac{1}{29} (x - 1)

29y-87=-(x-1)
29y+x-88=0

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