Question

what is the focus of the parabola given by the equation y=x^2-2x-3?

a. (4,-15/4)
b. (1,4)
c. (1,-4)
d. (1,15/4)
e. (1,-15/4)

Answer 1

If you need the focus, you'll have to complete the square, that way you can get it into the form
\[(x-h)^{2} = 4p(y-k)\]

Answer 2

i dont know how to do that @Concentrationalizing

Answer 3

YOu'll often have to complete the square in order to find the center and set up the proper equations for conic sections, so need to practice that.

So, if you have a general quadratic \(ax^{2} + bx + c\), then the first thing you would need to do is, if a is not 1, factor out a from the x^2 and x terms. Of course here a is 1, so we're fine.

Then you want to take half of b then square it. Our b value is -2, so (-2/2)^2 = (-1)^2 = 1. This value is added and subtracted at the same time from our expression. So if I do that, I have \(x^{2} -2x +1 -1 -3\)

After adding the 1 (I only subtracted 1 in order to keep the equation balance), I have a perfect square quadratic which can be factored into \((x-b/2)^{2}\). So this means I have this now:
\((x-1)^{2} -4\)
I combined the -1 and -3 to get the 4 there. So our original equation is equivalent to
\(y = (x-1)^{2} -4\)

Does that make any sense?