### 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.
That means completing the square for both x and y terms.
First, I'll group the x terms and y terms together and move the constant to the other side.
I get \[16x^2+16x-4y^2-20y=85.\]
Then, I'll factor out the coefficients of the squared terms, so I can complete the square inside the parentheses.
I get \[16(x^2+x)-4(y^2+5y)=85.\]
To complete the square, I need to add half of the linear coefficient squared to both sides.
For the x term, that's \(\frac{1}{2}\)^2, which is \(\frac{1}{4}\).
For the y term, that's \(\frac{5}{2}\)^2, which is \(\frac{25}{4}\).
Since I'm adding these inside the parentheses, I need to multiply them by the factors outside to keep the equation balanced.
So I get \[16(x^2+x+\frac{1}{4})-4(y^2+5y+\frac{25}{4})=85+16(\frac{1}{4})-4(\frac{25}{4}).\]
Simplifying, I get \[16(x+\frac{1}{2})^2-4(y+\frac{5}{2})^2=85+4-25.\]
Simplifying further, I get \[16(x+\frac{1}{2})^2-4(y+\frac{5}{2})^2=64.\]
Dividing both sides 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 with center at \(-\frac{1}{2},-\frac{5}{2}\), a horizontal transverse axis, and a and b equal to 2 and 4, respectively.
The vertices are the points on the transverse axis that are a units away from the center.
Since a is 2, and the transverse axis is horizontal, the vertices have the same y-coordinate as the center, which is -\(\frac{5}{2}\), and their x-coordinates are 2 units away from -\(\frac{1}{2}\).
That means the vertices are \(\left(-\frac{1}{2}+2,-\frac{5}{2}\right)\) and \(\left(-\frac{1}{2}-2,-\frac{5}{2}\right)\), or \(\left(\frac{3}{2},-\frac{5}{2}\right)\) and \(\left(-\frac{5}{2},-\frac{5}{2}\right)\).