# Thread: Quick tangent problem

1. ## Quick tangent problem

The tangent to the function $y=x^3-6x^2+8x$ at point $(3,-3)$intersects the curve at another point. How would I find that point?

What I did was: $f'(3)=\frac{y+3}{x-3}$

Where am I going wrong?

2. You have put "solved" on this. What solution did you get? Yes, what you did, saying that the slope of the line from (3,-3) to (x,y) must be the same as the derivative at 3, is correct. Although you shouldn't use "f" where only "y" was used before. Other than that, you are not going wrong at all.

3. Originally Posted by HallsofIvy
You have put "solved" on this. What solution did you get? Yes, what you did, saying that the slope of the line from (3,-3) to (x,y) must be the same as the derivative at 3, is correct. Although you shouldn't use "f" where only "y" was used before. Other than that, you are not going wrong at all.
Thanks for the pointers! And yes, I was able to figure it out. I set the equation of the tangent (using the formula above) with the equation of the function to find where the other intersection point was.