<s>[INST] Finding Asymptotes of Hyperbolas To find a asymptote its either b2/a2 or a2/b2 depending on the way the equation is written.
With the problem 
$$\frac{(x+1)^2}{16} - \frac{(y-2)^2}{9} = 1$$
The solutions the sheet I have is giving me is $3/4x - 3/4$ and $3/4 x + 5/4$
I thought it was just supposed to be $\pm 3/4x$. 
 [/INST] For the hyperbola $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$$
The asymptotes are $y - k = \pm \dfrac{b}{a}(x - h)$. 
You could leave your answer as $y - 2 = \pm \dfrac{3}{4}(x + 1)$, or write two separate equations. 

Edit... If you do write separate equations, you'll have
$y - 2 = \dfrac{3}{4}(x + 1)$ and $y - 2 = - \dfrac{3}{4}(x + 1)$, which are, in "slope-intercept form":
$y = \dfrac{3}{4} x + \dfrac{11}{4}$ and $y = - \dfrac{3}{4}x + \dfrac{5}{4}$
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