Help~ Don't remember what the point of tengency is.

[baby_garfield]

Here is the problem.

f(x)=x^3-18x^2=105x-146 and 2 is a real zero.

now, through the point (2,0), draw a line that is tangent to the graph at a point to the right of the low point at x=7. find the slope of the tangent line using x=2 and the point of tangency.

and here is the graph.

 

[HallsofIvy]

f(x)=x^3-18x^2=105x-146 and 2 is a real zero.

I assume that is f(x)= x³- 18x²+ 105 x- 146.
You want to find a line that passes through (2, 0) and is tangent to the graph at some (x,y) where x> 7.

Okay: any line through (2, 0) can be written as y= m(x-2) where m is the slope.

If the line is tangent to the graph at, say x₁ , then it must satisfy the equation at that point: we must have m(x₁- 2)= x₁³- 18x₁²+ 105 x₁- 146.
It also must have the same slope there so m= f '(x₁)= 3x₁²- 36x₁+ 105.

Solve those two equations for m and x₁.