# `y = x^3 - 3x, (2,2)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c)...

`y = x^3 - 3x, (2,2)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.

*print*Print*list*Cite

**a) **So we are given: `y=x^3 - 3x`

If we derive this function, then we will have the following:

`y'=3x^2 -3`

This will show us what the slope of the original line would be at any x value that we plug in. So if we want to know the equation of the line tangent to the graph at (2,2), then we have to plug in that x value:

`f'(2)=3(2)^2 -3=9`

So the slope of the graph at (2,2) should be 9, or 9/1

Since we have to have "m" and "b" for the slope-intercept(y=mx+b) form of an equation, we just need to find "b" now. We can do that by plugging in the slope (m), x value, and y value that we have now:

`2=(9)2+b`

`b=-16`

So the equation of the line should be:

`y=9x-16`