0
$\begingroup$

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

enter image description here

$\endgroup$
0
$\begingroup$

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 $2<k<4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.