Question

Find two distinct points on the curve y=x³+x²-x+1 that have parallel tangent lines

Answer 1

Given curve is $y=x^3+x^2-x+1.$

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

Let the two points which have parallel tangents lines be $(x_1,y_1)$ and $(x_2,y_2).$

Therefore, slope of both tangents must be equal.

$3{x_1}^2+2{x_1}-1=3{x_2}^2+2{x_2}-1\\ \Rightarrow 3({x_1}^2-{x_2}^2)+2({x_1}-{x_2})=0\\ \Rightarrow ({x_1}-{x_2})(3({x_1}+{x_2})+2)=0\\ \therefore x_1=x_2 \ or \ x_1+x_2=-\frac{2}{3}$

But, we know that $x_1\neq x_2$ because than both the points will be same.

So, $\ x_1+x_2=-\frac{2}{3}$

Hence, we can conclude that every two points whose sum of abscissa is $-\frac{2}{3}$ will satisfy this condition.