Question

what is the asympote of g(x)=1/(x-3)-4

Answer 1

simplify 12/x-13/^2

Answer 2

$$\frac{1}{x-3}+4$$Think about what happens to our terms as we let \(x\) get bigger, i.e. \(x\to\infty\). Do you notice a change in \(\frac1{x-3}\)? The denominator will grow while the numerator stays \(1\), and we can see that this results in the fraction becoming smaller and smaller, to \(0\); for large \(x\), it is actually so close to \(0\) that it's negligible for most uses. We also know that \(\frac1{x-3}\), as \(x\to3\), will have a smaller and smaller denominator; in fact, the function will "blow-up" so to speak. We know \(4\) doesn't depend on \(x\) and will always be \(4\), too.

Hey, wait, don't those sound like asymptotes? Right, \(\frac1{x-3}+4\) has a horizontal asymptote at \(y=4\) because \(\frac1{x-3}\to0\) as we let \(x\to\pm\infty\), and the \(4\) does not change -- so \((\frac1{x-3}+4)\to4\) as \(x\to\pm\infty\).
We also know that \(\frac1{x-3}\to\infty\) as we let \(x\to3\).