## Calculus 10th Edition

$f(x)$ is graph f
$f(x) = \frac{2x^2}{x^2+2} = \frac{2x^2 + 4 - 4}{x^2+2} = 2 - \frac{4}{x^2+2}$. $\frac{4}{x^2 + 2}$ never equals $0$ so $f(x)$ has a horizontal asymptote at $y = 2$. At $x=0$, $y = 0$ so that eliminates the options to c and f. At $x = -2$, $y$ is positive so this eliminates the options down to f.