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 $\pm 3/4x$. 
 A: 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 $y = - \dfrac{3}{4}x + \dfrac{5}{4}$
A: Hint: Your hyperbola is the standard hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, moved one unit to the left and $2$ units up. So write down the asymptotes of $\frac{x^2}{16}-\frac{y^2}{9}=1$, and move them in the same way.
Added: The asymptotes of $\frac{x^2}{16}-\frac{y^2}{9}=1$ are, as you know, $\frac{y}{3}=\pm \frac{x}{4}$. So the asymptotes in your case are
$$\frac{y-2}{3}=\pm\frac{x+1}{4}.$$
The "plus" case simplifies to $y-2=\frac{3}{4}x+\frac{3}{4}$,  then to $y=\frac{3}{4}x+\frac{11}{4}$.
The "minus" case simplifies to  $y=-\frac{3}{4}x+\frac{5}{4}$.  
A: For hyperbola $(x+1)^2/16 - (y-2)^2/9 = 1$, the equation for the asymptotes is $(x+1)^2/16 - (y-2)^2/9 = 0$.  This can be factored into two linear equations, corresponding to two lines.  The center of your hyperbola is $(-1,2)$, so of course the two asymptotes go through that point.
