### 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 a hyperbola, I need to put it in standard form, which is \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,\] where (h,k) is the center and a and b are the semi-major and semi-minor axes, respectively.
To do that, I need to complete the square for both x and y terms, and then divide by the constant term to make it equal to 1.
For the x terms, I have 16x^2+16x, which I can factor out a 16 to get 16(x^2+x).
Then I need to add a term inside the parentheses that makes it a perfect square.
The term is \[(\frac{1}{2})^2=\frac{1}{4},\] since \[(x+\frac{1}{2})^2=x^2+x+\frac{1}{4}.\]
So I have 16(x^2+x+\frac{1}{4})-16(\frac{1}{4}).
For the y terms, I have -4y^2-20y, which I can factor out a -4 to get -4(y^2+5y).
Then I need to add a term inside the parentheses that makes it a perfect square.
The term is \[(\frac{5}{2})^2=\frac{25}{4},\] since \[(y+\frac{5}{2})^2=y^2+5y+\frac{25}{4}.\]
So I have -4(y^2+5y+\frac{25}{4})+4(\frac{25}{4}).
Putting these together, I have \[16(x^2+x+\frac{1}{4})-16(\frac{1}{4})-4(y^2+5y+\frac{25}{4})+4(\frac{25}{4})-85=0.\]
Simplifying, I have \[16(x+\frac{1}{2})^2-4-4(y+\frac{5}{2})^2+25-85=0.\]
Rearranging, I have \[16(x+\frac{1}{2})^2-4(y+\frac{5}{2})^2=64.\]
Dividing by 64, I have \[\frac{(x+\frac{1}{2})^2}{4}-\frac{(y+\frac{5}{2})^2}{16}=1.\]
This is now in standard form, so I can identify the center, (h,k) as \[(-\frac{1}{2},-\frac{5}{2}).\]
I can also identify a and b as \[a=2 \text{ and } b=4.\]
The vertices are the points on the hyperbola that are a units away from the center along the transverse axis, which is horizontal in this case.
So the vertices are \[(-\frac{1}{2}\pm 2,-\frac{5}{2}),\] which are \[(-\frac{5}{2},-\frac{5}{2}) \text{ and } (\frac{3}{2},-\frac{5}{2}).\]
Either of these coordinates is a valid answer.