Find the points on the curve $y = x^{3} - 2 x^{2} - x$ , where the tangents are parallel to $3 x - y + 1 = 0$ .

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Solution

$\begin{matrix} y = x 3 - 2 x 2 - x . . . (1) l e t p o i n t s c u r v e o s (x_{1}, y_{1}) d i f f e r e n t i a t e w . r . t . x \frac{d y}{d x} = \frac{d (x 3 - 2 x 2 - x)}{d x} \frac{d y}{d x} = 3 x^{2} - 4 x - 1 s l o p e = \frac{d y}{d x} 3 x - y + 1 = 0.... (1) y = 3 x + 1 s l o p = 3 w h i c h i s p a r a l l e l t h e n 3 x^{2} - 4 x - 1 = 3 3 x^{2} - 4 x - 4 = 0 3 x^{2} - 6 x + 2 x - 4 = 0 3 x (x - 2) + 2 (x - 2) = 0 (3 x + 2) (x - 2) = 0 x = - \frac{2}{3} . o r x = 2 p u t t i n g x = - \frac{2}{3} o r x = 2 i n e q u a t i n (1) y = - \frac{8}{27} - \frac{8}{9} + \frac{2}{3} y = - \frac{14}{27} a n d y = - 2 t h e r e q p o i n t s a r e (- \frac{2}{3}, - \frac{14}{27}) a n d (2, - 2) \end{matrix}$

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Similar questions

y = x^{3} - 2 x^{2} - x

y = 3 x - 2

y = x^{3} - 2 x^{2} - x

y = 3 x - 2

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