Collinear Intersections

Subtracting the equation of the first line from the equation of the other, we get:

$x^{3} - 6x + 3 = 0 = f(x)$

The polynomial must have three real roots due to three intersections. Let the roots be named $a, b, c$ .

Hence, $f(a) = f(b) = f(c) = 0$

Consider the polynomial $g(x) = x^{3} + 2x - 1$

$g(x) = (x^{3} - 6x + 3) + (8x - 4)$

$g(x) = f(x) + (8x - 4)$

Hence, $g(a) = f(a) + (8a -4)$

Therefore, $g(a) = 8a - 4$

Therefore, in a similar manner,

$g(a) = 8a - 4$

$g(b) = 8b - 4$

$g(c) = 8c - 4$

Consider the intersection points. The graph intersects at $a, b, c$ . So we need to find the slope by picking two out of three of them. This time we will choose $a$ and $b$ .

Slope = $\huge\frac{\Delta y }{\Delta x} = \frac{g(b) - g(a)}{b - a} = \frac{(8b-4) - (8a-4)}{b-a} = \frac{8(b-a)}{b-a} = \boxed{8}$

$For~point~of~intersection~x,~and~y~values~will~be~the~same.\\ \implies~2x^3-4x+2=x^3+2x-1.~~\\ \therefore~x^3-6x+3=0.~solving~cubic,\\ x= -2.6691,~~~~~~.5240,~~~~~~2.1451.\\ Using~y=x^3+2x-1, ~~we~get.\\ x_{-2.6691}=-25.3531,~~~~x_{2.1451}=13.1608.\\ Slope=\dfrac{13.1608-(-25.3531)}{(2.1451-(-2.6691) }=\huge~~\color{#D61F06}{8.00006}.\\$