# Why do we only solve for $0$ on the numerator of this fraction?

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

• $\frac{a}{b}=0$ 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. – JMoravitz Oct 1 '19 at 1:24
• @JMoravitz would it be true to say that a fraction can only = 0 if numerator = 0? – Outsider Oct 1 '19 at 1:27
• Yes, again with the caveat that the denominator must be defined and nonzero there as well. – 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.