### Instruction: Find the coordinates of either of the vertices of the hyperbola \[16x^2+16x-4y^2-20y-85=0.\](Enter your answer as an ordered pair. Enter the coordinates of one of the vertices, not both.).### Response: To find the vertices of the hyperbola, I need to put it in standard form, which is either \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] or \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\] depending on whether the hyperbola opens horizontally or vertically.
To do that, I need to complete the square for both x and y terms.
For the x terms, I have \[16x^2+16x=16(x^2+x)=16(x^2+x+\frac{1}{4})-4=16(x+\frac{1}{2})^2-4.\]
For the y terms, I have \[-4y^2-20y=-4(y^2+5y)=-4(y^2+5y+\frac{25}{4})+25=-4(y+\frac{5}{2})^2+25.\]
So, the equation becomes \[16(x+\frac{1}{2})^2-4-4(y+\frac{5}{2})^2+25-85=0\] or \[16(x+\frac{1}{2})^2-4(y+\frac{5}{2})^2=64.\]
Dividing by 64, I get \[\frac{(x+\frac{1}{2})^2}{4}-\frac{(y+\frac{5}{2})^2}{16}=1.\]
This is the standard form of a hyperbola that opens horizontally, with center at \(-\frac{1}{2},-\frac{5}{2}\) and \(a=2\) and \(b=4\).
The vertices are at \((h\pm a,k)\), so they are \((-\frac{1}{2}\pm 2,-\frac{5}{2})\).
I can choose either one, so I'll pick \((\frac{3}{2},-\frac{5}{2})\).