how do you determine the best fit for the h(t)-axis (y-axis)

Question

h(t)=-16t^2+(29.5)t+(5) ,

-29.5 over 2(-16) = -29.5 over -32 = .92 ,

h(t) = 18.60

Answer 1

h(t)=-16(t²-29.5t/16)+5,

h(t)=-16(t²-29.5t/16+(29.5/32)²-(29.5/32)²)+5,

h(t)=-16(t-29.5/32)²+18.60 (approx).

When t=29.5/32=0.92 (approx), h(t) is maximum at 18.60.

The vertical scale must extend to at least 18.60. We also need the t intercepts, that is, when h(t)=0.

(t-29.5/32)²=18.60/16, t=0.92±1.08, t=-0.16, 2.

The intercept for h(t) is h(0)=5.

The width of the t-axis should be at least 2.16 to accommodate -0.16 and +2. So the horizontal and vertical axes have quite different scales. In round figures then the vertical axis should be about 20 units while the horizontal axis is only about 3 or 4 units with the vertical axis positioned close to the left. Mark the intercept 5 on the vertical axis.