Question

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

Answer 1

In the above question, we are given a curve as $y = {x^3} - 2{x^2} - x$ and a line whose equation is given as $y = 3x - 2$ . We have to find the point on the given curve such that the tangent to the curve $y = {x^3} - 2{x^2} - x$ is parallel to the given line $y = 3x - 2$ . We know that, for any two straight lines to be parallel, their slopes must be equal. So we have to find the slopes of the given straight line and the tangent to the given curve.

Complete step by step answer:
Given curve is, $y = {x^3} - 2{x^2} - x$
We know that the first derivative of a curve determines the slope of the tangent to that curve. Therefore slope of the given curve can be written as,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^3} - 2{x^2} - x} \right)$
That gives us the slope of the tangent as,
$\Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} - 4x - 1$
Now the equation of the given line is,
$\Rightarrow y = 3x - 2$

Comparing above equation with standard equation $y = mx + c$ we get,
$\Rightarrow m = 3$
Since, the tangent and the line are parallel, therefore their slopes must be equal.
Therefore, we can write
$\Rightarrow \dfrac{{dy}}{{dx}} = m$
That gives,
$\Rightarrow 3{x^2} - 4x - 1 = 3$
Subtracting $3$form both the sides, we get
$\Rightarrow 3{x^2} - 4x - 4 = 0$
Now the roots of the above equation can be written as,
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

That gives us,
$\Rightarrow x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 3 \right)\left( { - 4} \right)} }}{{2\left( 3 \right)}}$
Solving, we get
$\Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 48} }}{6}$
$\Rightarrow x = \dfrac{{4 \pm \sqrt {64} }}{6}$
Removing the square root, we have
$\Rightarrow x = \dfrac{{4 \pm 8}}{6}$
That gives us,
$\Rightarrow x = \dfrac{{12}}{6},\dfrac{{ - 4}}{6}$
Hence,
$\Rightarrow x = 2, - \dfrac{2}{3}$

Now, putting $x = 2, - \dfrac{2}{3}$ in the equation of the given curve, we get
$\Rightarrow y = {2^3} - 2 \cdot {\left( 2 \right)^2} - 2$
$\Rightarrow y = - 2$
And,
$\Rightarrow y = {\left( { - \dfrac{2}{3}} \right)^3} - 2 \cdot {\left( { - \dfrac{2}{3}} \right)^2} - \left( { - \dfrac{2}{3}} \right)$
$\Rightarrow y = - \dfrac{8}{{27}} - \dfrac{8}{9} + \dfrac{2}{3}$
That gives us,
$\Rightarrow y = \dfrac{{ - 8 - 24 + 18}}{{27}}$
Hence,
$\therefore y = - \dfrac{{14}}{{27}}$

Therefore, the tangents to the curve $y = {x^3} - 2{x^2} - x$ at the points $\left( {2, - 2} \right)$ and $\left( { - \dfrac{2}{3}, - \dfrac{{14}}{{27}}} \right)$ are parallel to the straight line $y = 3x - 2$.

Note: An algebraic equation of the second order is called a quadratic equation. A quadratic equation has only two roots. If a quadratic equation is given in the form of $a{x^2} + bx + c = 0$ , then the two roots of this equation are given by the formula,
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where the term $\sqrt {{b^2} - 4ac}$ is called the discriminant and two roots are:
-Real and distinct if $\sqrt {{b^2} - 4ac} > 0$
-Real and equal if $\sqrt {{b^2} - 4ac} = 0$
-Imaginary and distinct if $\sqrt {{b^2} - 4ac} < 0$