Answer to Question #309165 in Calculus for NURUL

Question

Find the coordinates of the points on the curve 𝑦 = 3𝑥³ − 2𝑥² − 12𝑥 + 2 where

the normal is parallel to the line 𝑦 =− 𝑥 + 1.

Answer 1

Find the coordinates of the points on the curve 𝑦 = 3𝑥³ − 2𝑥² − 12𝑥 + 2 where

the normal is parallel to the line 𝑦 =− 𝑥 + 1.

The derivative

$y'=9x^2-4x-12$

The slope of the normal is -1, hence the slope of the tangent line is $-\frac{1}{-1}=1$ .

Then

$y'=-1\Rightarrow 9x^2-4x-12=1\Rightarrow 9x^2-4x-13=0\Rightarrow x\in \{-1,\frac{13}{9}\}$

The points are

$(-1,9), (\frac{13}{9},-\frac{2543}{243})$