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

Question

Sorry This is the correct question

Answer 1

Why not go back to your previous version and make the necessary changes ??

Once you find the x's for both P and Q, (one of the solutions has to be x = -1)
sub the other x into the equation to find point Q

You know the slope is the same for both tangents and you know that slope
so find the equation of the tangent at Q

Now use the formula for the distance between a point and a line to find your 16√2 answer

Answer 2

I found the x for Q to be 3 and subbing x into the equation found y-coordinate as -20

Q(3,-20)
P(-1,8)

However using the distance equation I get 20root2 not 16root2. Can you check whether my working is right?

Answer 3

your other point is correct
so your equation of the other tangent is
y = x + b, remember the slope of both tangents is 1
plug in point (3,-20)
-20 = 3 + b
b = -23
the other tangent is y = x - 23 or x - y - 23 = 0

distance from (-1,8)
= |-1 - 8 - 23| / √(1^2 + (-1)^2 )
= 32/√2
= 32/√2 * √2/√2
= 32√2/2
= 16√2

Answer 4

Thanks, I calculated both tangents but was stuck on the next part. Thanks for explanation