I have this function where I need to find the $x$-intercepts

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I see they are $x = 0, 4$. And know that this is found by doing: $x=0, (x-4) = 0$ [Solve].

But why is this only done for the numerator?

  • 1
    $\begingroup$ $\frac{a}{b}=0$ if and only if $a=0$ (and $b$ is not zero, as it would be undefined in that case). In your case $\frac{x(x-4)}{(x-1)(x-5)(x+2)}=0$ if and only if the numerator $x(x-4)$ is zero. The denominator equalling zero won't give the intercepts, but rather the locations of the vertical asymptotes or poles. $\endgroup$ – JMoravitz Oct 1 '19 at 1:24
  • $\begingroup$ @JMoravitz would it be true to say that a fraction can only = 0 if numerator = 0? $\endgroup$ – Outsider Oct 1 '19 at 1:27
  • $\begingroup$ Yes, again with the caveat that the denominator must be defined and nonzero there as well. $\endgroup$ – JMoravitz Oct 1 '19 at 1:29

If you're asking why we set only the numerator equal to 0 and not the denominator, recall that $0$ divided by any nonzero number is $0$. Similarly, any number divided by $0$ is undefined.

Therefore in your case, the fraction is only equal to $0$ when its numerator is equal to $0$ and it's denominator is not equal to $0$, as having a denominator equal to $0$ (which is akin to dividing by $0$) would render the expression undefined.

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