Hyperbola and Rational Function

We can write $A$ in the form $\frac{(x-a)^2}{9}$ $-$ $\frac{(y-b)^2}{4}$ $=1$ since it is Horizontal and the slope of one of its asymptotes is $-\frac{2}{3}$ . This means that the slope of the other asymptote is $\frac{2}{3}$ . Because $A$ is moved to the point $(a,b)$ , we can apply the same transformation to the asymptotes, meaning the defined asymptote is $3y=-2x+(3b+2a)$ and the other one is $3y=2x+(3b-2a)$ .

We can perform Polynomial Division on $g(x)$ to get the quotient, or the oblique asymptote, as $y=$ $\frac{2}{3a}$ $x$ + $\frac{6a^2b-2ab}{3a^3}$ . There is no cross point since the remainder has to be a nonzero constant. Since the slope is $\frac{2}{3}$ , $a=1$ and the $y$ -intercepts have to be the same. As a result, $\frac{3b-2}{3}$ $=$ $\frac{4b}{3}$ and $b=-2$ . As a result, $5a+4b=5(1)+4(-2)=-3$ which is the final answer.