keeping in mind that the vertex is between the focus point and the directrix, in this cases we have the focus point above the directrix, meaning is a vertical parabola opening upwards, Check the picture below, which means the "x" is the squared variable.
now, the vertical distance from the focus point to the directrix is 
, which means the distance "p" is half that or 1/8, and is positive since it's opening upwards.
since the vertex is 1/8 above the directrix, that puts the vertex at 
, meaning the y-coordinate for the vertex is 2.
![\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertical%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Ck%2Bp%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7By%3Dk-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \begin{cases} h=-4\\ k=2\\ p=\frac{1}{8} \end{cases}\implies 4\left(\frac{1}{8} \right)(y-2)=[x-(-4)]^2\implies \cfrac{1}{2}(y-2)=(x+4)^2 \\\\\\ y-2=2(x+4)^2\implies \blacktriangleright y = 2(x+4)^2+2 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20h%3D-4%5C%5C%20k%3D2%5C%5C%20p%3D%5Cfrac%7B1%7D%7B8%7D%20%5Cend%7Bcases%7D%5Cimplies%204%5Cleft%28%5Cfrac%7B1%7D%7B8%7D%20%5Cright%29%28y-2%29%3D%5Bx-%28-4%29%5D%5E2%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28y-2%29%3D%28x%2B4%29%5E2%20%5C%5C%5C%5C%5C%5C%20y-2%3D2%28x%2B4%29%5E2%5Cimplies%20%5Cblacktriangleright%20y%20%3D%202%28x%2B4%29%5E2%2B2%20%5Cblacktriangleleft)
Answer: I would say the third one
Step-by-step explanation: I hope this helps if it is wrong i am so sorry I hope you have a great day of night stay safe :)
Answer:
The dimension of the play area written algebraically is y²/2 + 9y
Step-by-step explanation:
Let the be x and the width be y
If the length is to be 9 feet longer than half the width,
hence y/2 + 9 = x
hence dimension of the area = (y/2 + 9) × y = y²/2 + 9y
Answer:
HOLA BB COM O ESTAS DIME QUE TE AYUDO BB
Step-by-step explanation:
