Given $f(x)$'s graph, how can I obtain $g(x)$'s graph knowing $ (f(x) )^2 = g(x)$

Question

The equation $$ f(x) = \frac{x+2}{x-2}$$ has two asymptotes: $x=2$ and $y=1$. $f(x)$'s graph looks sort of like a distorted hyperbola. So, after figuring that out -- through limits or whatever --, how can I find what the graph of $g(x) = f(x)^2$ looks like?

Answer 1

Note that $$ f(x) = 1 + \frac{4}{x-2} $$ so it is a scaled and shifted hyperbola, as you correctly note, with vertex at $(2,1)$. Please see the graph below:

When you square the function, we get $$ g(x) = f(x)^2 = \left(1+\frac{4}{x-2}\right)^2 $$ on a very high level you expect the vertical asymptote at $x=2$ to stay, and the right piece of the hyperbola not to globally change behavior, but the lower-left half of the hyperbola should "reflect over the $y$-axis" (reflection is not exact here since shape will change, but high level it should approach $+\infty$ as $x \to 2^-$.)

Note that as $x \to -\infty$, you should have $g(x) \to 1-$ (as $f(x) \to 1-$ there, so you expect there would be a parabola-like bend from $1$ at $x\to-\infty$ and $+\infty$ at $x \to 2^-$, with some relative minimum somewhere in between, which you can find by solving $$ 0 = g'(x) = 2f(x)f'(x) \iff f'(x) = 0 \quad \text{or} \quad f(x)=0 $$

Indeed, here is the graph: