consider the line y= 10-2x amd the function y(x)=2x^3 - 3x^2 -14x +3. determine the coordinates of the point(s) where the line and the graph of the function are tangent.
THANK YOU!!!

Question

consider the line y= 10-2x amd the function y(x)=2x^3 - 3x^2 -14x +3. determine the coordinates of the point(s) where the line and the graph of the function are tangent.
THANK YOU!!!

tkhunny · Answer

The slope of the line is what?
The general expression for the slope of the tangent is what?

anonymous · Answer

slope of line =2
and i didnt get the second part

tkhunny · Answer

Have you considered the 1st Derivative, y'(x)?

anonymous · Answer

oh, that would be 6x^2 -6x-14, right?

tkhunny · Answer

Yes.  Where does y'(x) = -2?

BTW, the slope of the line is -2, not just 2.

anonymous · Answer

i'l try to work out the final answer and i'll post it

anonymous · Answer

i found$$0.5+(\sqrt{420}/12)$$ and  $$0.5-(\sqrt{420}/12)$$  not sure how to simplify the sqrt of 420/12

tkhunny · Answer

Oh, how I wish you had shown your work.  Something went seriously wrong.

You should be trying to solve: $6x^{2} - 6x - 14 = -2$

Is that where you started?

anonymous · Answer

yeah sry, i sent -2 to the other side of the equation instead of 2.

so,
D= 324
 $$0.5-( \sqrt{324}/12), 0.5+( \sqrt{324}/12)$$

tkhunny · Answer

$\sqrt{324} = 18$

anonymous · Answer

x=-1     x=2
am i supposed to use the -1?

tkhunny · Answer

Why would you ignore it?  There are two solutions.