Answer to Question #305328 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

$$y=3x^3-2x^2-12x+2$$

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

The slope of the normal is

$$slope=-\dfrac{1}{f'(x_0)}=-1$$

Then

$$9x_0^2-4x_0-12=1$$

$$9x_0^2-4x_0-13=0$$

$$x_0=\dfrac{13}{9}\ or\ x_0=-1$$

$x_0=\dfrac{13}{9}$

$$y(\dfrac{13}{9})=3(\dfrac{13}{9})^3-2(\dfrac{13}{9})^2-12(\dfrac{13}{9})+2=-\dfrac{2543}{243}$$

$x_0=-1$

$$y(-1)=3(-1)^3-2(-1)^2-12(-1)+2=9$$

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