Question

Find point on curve $y=x^3-2x^2-x$ at which tangent lines are parallel to line $y=3x-2$

Answer 1

Equation of curve is $y=x^3-2x^2-x$ …………$\left(1\right)$

$\frac{dy}{dx}=3x^2-4x-1$

$\therefore$ At $(x_1,y_1)$ slope of tangent $\left(m_1\right)=3x_1^2-4x_1-1$ ……….$\left(1\right)$

Slope of line $y=3x-2$

$m_2=3$

Since, the tangent is parallel to line

$y=3x-2$

$m_1=m_2$

$\therefore 3x_1^2-4x_1-1=3$

$3x_1^2-4x_1-4=0$

$3x_1^2-6x_1+2x_1-4=0$

$3x_1(x_1-2)+2(x_1-2)=0$

$(x_1-2)(3x_1+2)=0$

$x_1=2$ or $x_1=-\frac{2}{3}$

When $x_1=2$

Substituting $x_1=2$ in equation $\left(1\right)$, we get

$y_1=(2{)}^3-2(2{)}^2=2$

$y_1=-2$

When, $x_1=\frac{-2}{3}$

$y_1={\left(\frac{-2}{3}\right)}^2-2\left(\frac{-2}{3}\right)-\frac{2}{3}$

$y_1=\frac{-14}{27}$

$\therefore$ Points on curve $y=x^3-2x^2-x$ at which tangent lines are parallel to line $y=3x-2$ are $(2,-2)$ and $(\frac{-2}{3},\frac{-14}{27})$.