Question

find the gradient of the curve y=2x^3-5x^2-x+1 at the pt (2,-5). find the x-coordinate of another pt on the curve where the tangent at that pt is parallel to the tangent at the pt (2,-5)

Answer 1

rather long question.....
dy/dx = 6x^2 - 10x - 1
at (2,-5) , dy/dx = 3

equation of tangent at that point:
y = 3x+b, with (2,-5) on it
-5 = 6+b
b = -11
y = 3x - 11

so now intersect: set y = y
2x^3 - 5x^2 - x + 1 = 3x - 11
2x^3 - 5x^2 -4x + 12 = 0
We already know that x=2 is a solution
so x-2 is a factor.
by synthetic division I found
2x^3 - 5x^2 -4x + 12 = (x-2)(2x^2 - x - 6)
= (x-2)(x-2)(2x+3)

we have a double root, which is the case where a tangent touches the curve, and another at
x = -3/2
y = ...... You do the arithmetic.