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.