# Homework Help: Finding points on the graph

1. Sep 13, 2011

### brettevan

I have absolutely no idea how to solve this problem and I would really appreciate same help.

For the function, find the point(s) on the graph a which the tangent line has slope 5

y=1/3x^3-2x^2+8x+14

Any help with this would br greatly appreciated

2. Sep 13, 2011

### Dick

Do you know what a derivative is and what it might have to do with a slope? You really have to make a better start than that.

3. Sep 14, 2011

### Ray Vickson

Use brackets. I really cannot figure out whether you mean $$y = \frac{1}{3} x^3 - 2 x^2 + 8x + 14$$ or $$\frac{1}{3x^3} - 2x^2 + 8x + 14$$ or $$y =\frac{1}{3x^3 - 2x^2 + 8x + 14} .$$ Strictly according to standard rules, what you wrote is the first of these three possibilities. If you don't want to use Latex you can just write (1/3)x^3 - 2x^2 + ... or 1/(3x^3) - 2x^2 + ... or 1/(3x^3 - 2x^2 + ...)

RGV