whats the vertex of 4y^2+4y-16x+13=0?

Answer 1

rearrange, then complete the square ....
16x = 4y^2 + 4y + 13
4x = y^2 + y + 1/4 - 1/4 + 13/4
4x = (y + 1/2)^2 -1/4 + 13/4
4x = (y+1/2)^2 - 3
x = (1/4)(y + 1/2)^2 - 3/4

can you read the vertex from that?

Answer 2

so is it (-3/4,-1/2)?

Answer 3

yes, good for you to notice that you had to read the x and y in inverse form.