Find the vertex, focus, and directrix. y = 1/24(x+1)² - 3.

Answers

Answer 1
Answer: y = (1)/(24)(x+1)^2 - 3\n\ny+3 =(1)/(24)(x+1)^2\ \ / *24\n\n (x+1)^2 = 24(y+3)

This   is  an  equation  of  a  parabola  that  opens  upwards.

Its \ standard \ form: \n(x-h)^2=4p(y-k)\n (h,k)=(x,y) \ coordinates \ of \ the \ vertex\n\ (h,k)=(-1,-3) \n\naxis \ of \ symmetry: \ x= -1\n \n4p=24\ \ /:4\np=6

focus:(h,k+p)=(-1,-3+6)=(-1,3) \n \ndirectrix: \ y=k-p=-3-6=-9


Answer 2
Answer: the\ equation\ in\ the\ form\ (x-h)^2=4p(y-k)\ is \ a\ parabola\nwith\ a\ vertex\ at\ \ (h,\ k), \na\ focus\ at\ \ (h,k+p)\n\ and\ a\ directrix\ \ y = k - p \n\n y = 1/24(x+1)^2 - 3\ \ \ \ \Rightarrow\ \ \ y+3 = 1/24(x+1)^2\ /\cdot24\n\n 24\cdot(y+3)=(x+1)^2\n\n(x+1)^2=4p(y+3)\ \ \Rightarrow\ \ 4p=24\ \ \Rightarrow\ \ p=6\ \ \ and\ \ \ h=-1,\ k=-3\n\nthe\ vertex:\ \ \ (h;\ k)=(-1;\ -3)\n\nthe\ focus:\ \ \ (h;\ k+p)=(-1;\ -3+6)=(-1;\ 3)\n\nthe\ directrix:\ \ \ y=k-p\ \ \ \Rightarrow\ \ \ y=-3-6=-9

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Answers

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Answers

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Answers

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