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anonymous
 one year ago
what is the focus of the parabola given by the equation y=x^22x3?
a. (4,15/4)
b. (1,4)
c. (1,4)
d. (1,15/4)
e. (1,15/4)
anonymous
 one year ago
what is the focus of the parabola given by the equation y=x^22x3? a. (4,15/4) b. (1,4) c. (1,4) d. (1,15/4) e. (1,15/4)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you need the focus, you'll have to complete the square, that way you can get it into the form \[(xh)^{2} = 4p(yk)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how to do that @Concentrationalizing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YOu'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 \((xb/2)^{2}\). So this means I have this now: \((x1)^{2} 4\) I combined the 1 and 3 to get the 4 there. So our original equation is equivalent to \(y = (x1)^{2} 4\) Does that make any sense?
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