Question 1194922
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Red = {{{ 4(sqrt(x+2)) }}} = f(x)
Blue = normal
Green = tangent

   *[illustration f_x_with_tan_n_norm].



Tangent line:
dy/dx = {{{ (1/2)(4)(x+2)^(-1/2) }}} = {{{ 2/sqrt(x+2) }}}

This is the slope of the tangent line.  Using the fact that the tangent meets the function f(x) at (7,12) let's us find the y-intercept:

    At x=7, dy/dx = {{{ 2/sqrt(7+2) = 2/3 }}}

     Now,   y = (2/3)x + b
            12 = (2/3)*7 + b   ==>  b = 22/3

     You can now write  y = (2/3)x + 22/3  for the tangent line
     Set y=0 to find the x-axis crossing:
                 0  = (2/3)x + 22/3
      Solving for x gives  x = -11   

   From this, we conclude M is at (-11,0) 
      

Normal line:
 The slope of the normal line will be  -1/(slope of tangent).  We are only interested in (7,12), so slope of normal is  -(1/(2/3)) = -3/2

       You can now write  y = (-3/2)x + b  for the normal line
      Use the fact that the normal line also passes through (7,12) to find 
      the y-intercept of the normal:
                   12 = (-3/2)(7) + b   ==>  solve for b ==>  b = 45/2
 
       We can now write   y = (-3/2)x + 45/2   for the normal line

      Now find the x-axis crossing by setting y=0:

             0 = (-3/2)x + 45/2  ==>  solve for x ==>  x = 15

               So N is at  (15, 0)

Area of PMN:
P is at (7,12)  so you can calculate  |PN| and |PM|   ... and then use the triangle area formula   A = (1/2)|PN| * |PM|

I got 156.00 sq units when I did these calculations.