Differential calculus
Differential Calculus - Derivatives and differentials
Determining the lines tangent to the graph of a function from a point outside the function
Determining the lines tangent to the graph of a function from a point outside the function, example
Determining the lines tangent to the graph of a function from a point outside the function
Lines tangent to the graph of a function y = f (x) from a given point (x1, y1) outside the function are defined by two points they pass through, the given point (x1, y1) and the point of tangency (x0, y0).
To find the coordinates x0 and y0 of the points of tangency we should solve the following two equations,
(1)   y0 = f (x0)
(2)   y1 -  y0 =  f ' (x0) · (x1 - x0)
as the system of the two equations with the two unknowns x0 and y0.
The first equation states that the point of tangency must lie on the given function while, at the same time, its coordinates x0 and y0 must satisfy the equation of the tangent line (2), since the point of tangency is the common point of the tangent and the curve.
Thus for example, if given is the quadratic function   f (x) =  ax2 + bx + c and the point  P1(x1, y1) t hen, applying the above system of equations we can calculate the coordinates of the points of tangency                 P0 (x0, y0)  directly from the formulas,
 Example: Find equations of the lines tangent to the function  f (x) = - x2 + 2x + 4 drawn from the point
(3, 2).
 Solution:  To find the coordinates x0 and y0 of the points of tangency we should solve the system of equations. Substitute given quantities, the point (3, 2) and  f (x) = - x2 + 2x + 4  into equations,
Let calculate the slope  f ' (x0) of the tangent line, the equation (2).
Then substitute the slope and the y0 into second equation to get the points of tangency,
 Therefore, the points of tangency are (2, 4) and (4, -4). To write the equations of tangents t1 and t2  we calculate their slopes m1 and m2 from f ' (x0) = - 2x0 + 2. Thus,     m1 = f ' (2) = - 2 · 2 + 2 = - 2,    m1 = - 2, and      m2 = f ' (4) = - 2 · 4 + 2 = - 6,    m2 = - 6. The tangent t1 through the point (2, 4) and m1 = - 2, and the tangent t2 through the point (4, -4) and m2 = - 6, t1 ::   y = - 2x + 8  and  t2 ::   y = - 6x + 20.
Check the coordinates of the points of tangency by plugging the corresponding parameters into the above formulas.
Functions contents F