### Instruction: The hyperbola \[\frac{(x-3)^2}{5^2} - \frac{(y+1)^2}{4^2} = 1\]has two asymptotes, one with positive slope and one with negative slope. Compute the $x-$intercept of the one with positive slope. (Enter your answer as an ordered pair.).### Response: I recall that the asymptotes of a hyperbola are given by the equations \[y = \pm \frac{b}{a} (x-h) + k,\] where $(h, k)$ is the center of the hyperbola and $a$ and $b$ are the semi-major and semi-minor axes respectively.
In this case, the center is $(3, -1)$, $a = 5$, and $b = 4$.
The asymptote with positive slope has the equation \[y = \frac{4}{5} (x-3) - 1.\]
To find the $x-$intercept, I need to set $y$ to zero and solve for $x$.
Doing so, I get \[0 = \frac{4}{5} (x-3) - 1 \implies x = \frac{17}{4}.\]
Therefore, the $x-$intercept is $(\frac{17}{4}, 0)$.