The tangent to the function $\displaystyle y=x^3-6x^2+8x$ at point $\displaystyle (3,-3) $intersects the curve at another point. How would I find that point?
What I did was: $\displaystyle f'(3)=\frac{y+3}{x-3}$
Where am I going wrong?
The tangent to the function $\displaystyle y=x^3-6x^2+8x$ at point $\displaystyle (3,-3) $intersects the curve at another point. How would I find that point?
What I did was: $\displaystyle f'(3)=\frac{y+3}{x-3}$
Where am I going wrong?
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.