Sketching Logs with Quadratic Terms

Question

$\log(x^2+1) = y$

asymptote at $x^2+1 > 0$ and so there is no asymptote

$x$ and $y$ intercept at $(0,0)$

How do you know that the function goes both directions, and has a dip in the middle?

graph is shown below:

Answer 1

Let's do some heuristical reasoning: I assume you have some knowledge about the $log(x)$ function: this function is defined for $x \gt 0$ and keeps on growing when $x$ get larger and larger. So certainly your function $log(x^2+1)$ keeps growing in both directions, when $x$ gets very large and positive or very large negative. So asymptotes, forget it!

Also $log(x)$ is negative only when $0 \lt x \lt 1$, implying that $log(x^2+1)$ is always positive (for all $x$, $x^2+1 \gt 1$), that is, does not lie under the $X$-axis. Finally, it does not matter whether you plug in $x$ or $-x$ in your function, it yields the same value $y$. In other words, the graph must be symmetric around the $Y$-axis (in sophisticated terminology: it is an even function). You already observed that $(0,0)$ is a point on the graph.

So, now your are ready to roughly sketch it. Observe that up to this point we have not done any serious calculation and only extrapolated our knowledge from the $log(x)$ function to yours. But if you want to get more exact, you can calculate more points on the graph or take its derivative ($y'=\frac{2x}{x^2+1}$), calculate its minimum, whatever etc..