This is a priority! the equation of a parabola is 1/16 (y+3)^=x+4 what are the coordinates of the focus?

To solve this problem you must apply the proccedure shown below:

 1. You have the following equation of a parabola:

 1/16(y+3)^2=x+4 

 2. Rewriting the equation of the parabola, you have:

 (y+3)^2=16(x+4)

 3. The vertex of the parabola is:

 (h,k)=(-4,-3)

 3. Now, you must find the coefficient of the unsquared part, as below:

 4p=16
 p=16/4
 p=4

 4. As you can see, the y part is squared and p is positive. This means that the parabola that opens to the right.

 5. The focus of the parabola has to be 4 units to the right of the vertex. Therefore, the focus is:

 (0,-3)

 The answer is: (0,-3)

143 75 107 13 the answer

/-/////////looks tough ‍♂️