Question

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

Answer 1

$y = \frac{1}{24}(x+1)^2 - 3\\\\y+3 =\frac{1}{24}(x+1)^2\ \ / *24\\\\ (x+1)^2 = 24(y+3)$

This   is  an  equation  of  a  parabola  that  opens  upwards.

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

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

Answer 2

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