$\begingroup$

Consider the function$$f(x) = \frac{-4}{x-2}$$

To find the $x$-intercept, $y$ is set equal to $0$. However, in rational functions, the numerator is set equal to $0$. In the example above, since the numerator is constant, there is no $x$-intercept. Why is the numerator the only one taken into account when finding the $x$-intercept in rational functions?

$\endgroup$ 0

2 Answers

$\begingroup$

The $x$-intercepts are given by solving $f(x) =0$. In your case, this is $$ - \frac{4}{x-2} =0.$$ Multiply both sides by $x-2$, you get $-4 =0$, which is nonsense. Hence, $f(x)$ has no $x$-intercepts, and thus has a horizontal asymptote at $y=0$.

$\endgroup$ $\begingroup$

Well, the denominator must be taken into account as well. Consider $f(x)= \frac{x-1}{x-1}$. The numerator vanishes at $x=1$, but $f$ has no zeroes at all.

A quotient has a zero where the numerator vanishes and the denominator does not.

$\endgroup$ 2

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

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