## Calculus: Early Transcendentals 8th Edition

a. Find the vertical asymptotes $x = 0$ and $x = \frac{3}{2}$
$y = \frac{x^{2} + 1}{3x - 2x^{2}}$ 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 $x = \frac{3}{2}$.