y' = 3x^2 - 6x - 8
at (1,-8), y' = 3 - 6 - 8 = -11
So at Q, the slope of the tangent must also be -11
3x^2 - 6x -8 = -11
3x^2 - 6x + 3 = 0
x^2 - 2x + 1 = 0
(x-1)^2 = 0
x = 1
mmmhhh, expecting two values of x
y'' = 6x-6
= 0 for point of inflection
x = 1
if x = 1, y = 1-3-8+4 = -6
ahhh, found the problem.
you said that there was a tangent at (1, -8), but that point does not lie on the curve
There is a typo, or the question is bogus.
The curve y=x^3-3x^2-8x+4 has tangent L at point P (1,-8). Given that the Line M is parallel to L and is also a tangent to Q show that the shortest distance between L and M is 16 root 2
Thanks
1 answer
1 answer