Find the x‑ and y‑coordinates of the focus of the parabola that is represented by the equation below. Enter your answers as integers. y=0.125x2+4x-29 Focus

In form

4p (y-k) = (x-h) ^2

vertex = (h, k)

p is the horizontal distance from the vertex to the focus and directix

when p is positive, the focus is above the vertex

complete the square

y = (0.125x^2+4x) - 29

y=0.125 (x^2+32x) - 29

32/2=16, 16^2=256

add negative and positive inside

y=0.125 (x^2+32x+256-256) - 29

complete the square

y=0.125 ((x+16) ^2-256) - 29

distribute

y=0.125 (x+16) ^2-32-29

y=0.125 (x+16) ^2-61

add 61 to both sides

y+61=0.125 (x+16) ^2

divide both sides by 0.125

8 (y+61) = (x+16) ^2

4 (2) (y+61) = (x+16) ^2

4 (2) (y - (-61)) = (x - (-16)) ^2

4p (y-k) = (x-h) ^2

vertex at (-16,-61)

p is 2 which is positive so it s above

-61+2=-59

focus at (-16,-59)