Equation of a line that is tangent to f(x)

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The discussion focuses on finding the equation of a line that is tangent to the function f(x) = 4x - x² and passes through the point P(2, 7), which is not on the graph of f(x). To solve this, one must first compute the derivative f'(x) to determine the slope at the tangent point. The user attempts to use the line equation y = mx + b, substituting the slope derived from f'(x) and the coordinates of point P to find the correct tangent line equation.

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In order to find the equation of a line that is tangent to f(x) and goes through point P on f, you got to find the derivative of f(x) at P, but how would you go about solving a problem where you have to find the equation of a line tangent to f(x) that goes through point P, but P is NOT on the graph of f.
 
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Just to get my facts straight: you have a function [tex]f(x)[/tex], a point [tex]P[/tex] that is not on the graph of [tex]f(x)[/tex], and you want an equation of a line that

  1. passes through the point [tex]P[/tex], and
  2. is tangent to the graph of [tex]f(x)[/tex]

If this is correct, what else is stated in the problem - do you have a specific function [tex]f[/tex], at what point(s) is the line to be tangent, etc. Further, what have you tried?
 


f(x) = 4x-x2

Question: Find the equations of the lines that pass through P(2,7) and are tangent to the graph of f(x).

(P is not on f(x).)
Thats all the problem states.

Ive tried finding f'(x) and plugging f' into the Line equation y=mx+b.

y=(4-2x)x+b.

Then plugging in Point P.

7=(4-2x)2+b - I am not really sure if this is heading in the right direction.
 

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