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anonymous

  • one year ago

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)

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

  2. anonymous
    • one year ago
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    i dont know how to do that @Concentrationalizing

  3. anonymous
    • one year ago
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    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?

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