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
