"What is the equation of the normal line..." was answered by our community

Nevaeh Glass

Answered question

2022-04-10

What is the equation of the normal line of f(x)=(x-1)(x+4)(x-2) at x=3?

ysnlm8eut

Beginner2022-04-11Added 14 answers

Explanation:
before differentiating , distribute the brackets.
start with

(x - 1) (x + 4) = x^{2} + 3 x - 4

(x^{2} + 3 x - 4) (x - 2) = x^{3} - 2 x^{2} + 3 x^{2} - 6 x - 4 x + 8

f (x) = x^{3} + x^{2} - 10 x + 8

The derivative of f(x) is the gradient of the tangent and f'(3)
the value of the tangent.

f^{'} (x) = 3 x^{2} + 2 x - 10

and

f^{'} (3) = 3 {(3)}^{2} + 2 (3) - 10 = 27 + 6 - 10 = 23 = m

For 2 perpendicular lines , the product of their gradients
is -1
let

m_{1} =

gradient of normal
then

m \cdot m_{1} = - 1 \to 23 \cdot m_{1} = - 1 \Rightarrow m_{1} = - \frac{1}{23}

f (3) = (2) (7) (1) = 14 \Rightarrow (3, 14)

is normal point
equation of normal is

y - b = m (x - a) m = - \frac{1}{23}, (a, b) = (3, 14)

y - 14 = - \frac{1}{23} (x - 3)

(multiply by 23 to eliminate fraction )

23 y - 322 = - x + 3 \Rightarrow 23 y + x - 325 = 0

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