As long as f'(x) != 0, the normal line of f(x) at x can be written on the form y = ax + m, where a is the slope and m is the intercept with the y-axis (see further down what happens for the case f'(x) = 0). If we first find a, then we can find m through elimination.
We know that the normal line of f(x) is perpendicular to the tangent line of f(x) at x=3. Therefore, if we can find the slope of f(x) at x=3, then we can also find the slope of the normal line at x=3.
Denote the slope of the tangent line b. If a,b != 0, then a*b=-1 (see further down for explanation).
The slope of the tangent line at a value x is by definition f'(x), which we can compute using the product rule (or quotient rule) .
f'(x) = (1)/(x+4) - (x-3)/(x+4)^2 = (x+4)/(x+4)^2 - (x-3)/(x+4)^2 = ((x+4)- (x-3))/(x+4)^2 = 7/(x+4)^2.
Therefore b = f'(3) = 7/7^2 = 1/7, and we find the slope of the normal line by solving a*b=-1 for a, which gives that
a = -7.
Since the normal line passes through x=3, we know that the point (3,f(3)) must lie on the line. Evaluating f(3) = 0, and inserting in the equation for the normal line we get that
0 = a * 3 + m.
Inserting a = -7 and solving for m gives that
m = 21,
and we have found that the normal line is given by
y = -7x + 21.
Comment 1: In case f(x)=0, the tangent to f(x) is flat, which means that the normal line is vertical. Then we must use the general equation for the line
py + qx = r,
which, since the normal line is vertical must have p = 0. To find q and r, insert a pair of x and y-values that you know lie on the line, like we did earlier.
Comment 2: In case f(x) is not differentiable at x, then the function does not have a normal line there.
Comment 3: As for showing the formula a*b=-1, you need either geometry, trigonometry or linear algebra. Here are some proofs.
