use 4x^2-y^2-8x-4y+16 to do the following
write the equation in standard form,and located the centre and the vertices of this curve
also,find the domain and range of this curve
The equation hasn't been copied down as such so I am guessing as to what it is.
Group by variable and Factor:
$\displaystyle 4(x^2-2x) -(y^2 + 4y) - 16 = 0
$
Adding a 1 inside the parenthesis really means that you have added 4 to the left side. Also, in the group of y-terms, adding 4 inside, with a minus in the front, means you have subtracted 4 from the left side. So we do the same two things to the right side:
$\displaystyle 4 (x^2 - 2x + 1) - (y^2 + 4y + 4) = 16 + 4 - 4$
Complete the square, then divide both sides by 16, to get
$\displaystyle \frac{(x-1)^2}{4} - \frac{(y+2)^2}{16} = 1
$
Does this resemble a form of equation you're familiar with?
The rest should be straight-forward. Good luck!!