Tangent is when the derivative is zero. That is #4x - 1 = 0. x = 1/4# At x = -2, f' = -9, so the slope of the normal is 1/9. Since the line goes through #x=-2# its equation is #y = -1/9x + 2/9#
First we need to know the value of the function at #x = -2#
#f(-2) = 2*4 + 2 + 5 = 15#
So our point of interest is #(-2, 15)#.
Now we need to know the derivative of the function:
#f'(x) = 4x - 1#
And finally we'll need the value of the derivative at #x = -2#:
#f'(-2) = -9#
The number #-9# would be the slope of the line tangent (that is, parallel) to the curve at the point #(-2, 15)#. We need the line perpendicular (normal) to that line. A perpendicular line will a negative reciprocal slope. If #m_(||)# is the slope parallel to the function, then the slope normal to the function #m# will be:
#m = - 1/(m_(||))#
This means the slope of our line will be #1/9#. Knowing this we can proceed with solving for our line. We know it will be of the form #y = mx + b# and will pass through #(-2, 15)#, so:
#15 = (1/9)(-2) + b#
#15 + 2/9 = b#
#(135/9) + 2/9 = b#
#b = 137/9#
This means our line has the equation:
#y = 1/9x + 137/9#
