# finding the behavior of the asymptotes in a rational function

I'm having trouble understanding how to graph this function: $f(x) = \frac{x-2}{(x-4)(x+4)}$.

The part I undertand:

The x-intercept is (2,0) since x=2 makes the numerator zero. The y-intercept is (0,1/8) as $f(0) = 1/8$

The vertical asymptotes are -4 and 4 as these are the values for which the denominator of $f(x) = \frac{x-2}{(x-4)(x+4)}$ equals zero. The horizontal asymptote is y = 0 since the degree of the numerator is less than the degree of the denominator in $f(x) = \frac{x-2}{(x-4)(x+4)}$

The part I don't understand

In the image below, how do we know the third column of the table? i.e. how do we know the sign of f(x) is negative if we use a test value of 3.999 for as $x \to 4-$ Can you please explain the table

Let $k$ be some number 'slightly less' than $4$. If you plug it into the function you can consider the positivity of each of the factors. $$f(k) = \frac{k-2}{(k-4)(k+4)}\sim \frac{(+)}{(-)(+)}$$ So the overall function is negative at $k$ since $2<k<4$.