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)

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote