#### Answer

a. Find the vertical asymptotes
$ x = 0$ and $x = \frac{3}{2}$

#### Work Step by Step

$y = \frac{x^{2} + 1}{3x - 2x^{2}}$
Take out the common factor "$x$" in the denominator.
$y = \frac{x^{2} + 1}{x(3 - 2x)}$
Then set the denominator equal to zero and solve for x.
$x = 0$
$3-2x = 0$
$3 = 2x$
$x = \frac{3}{2}$
The vertical asymptotes are $x = 0$ and $x = \frac{3}{2}$.