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

  • #1

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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.
 

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  • #2
HallsofIvy
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baby garfield said:
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.
 

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