Answer:
Domain: [0, 12.75]
Range: [0, 590.9]
Step-by-step explanation:
Your function is missing variables.
I think you mean
h(t) = -4.922t^2 + 17.69t + 575
as the function.
We can find the time the object hits the ground. At that time, h(t) = 0, so we set the equation equal to zero and solve for t.
-4.922t^2 + 17.69t + 575 = 0
[tex] t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex]
[tex] t = \dfrac{-17.69 \pm \sqrt{(-17.69)^2 - 4(-4.922)(575)}}{2(-4.922)} [/tex]
[tex] t = \dfrac{-17.69 \pm \sqrt{11633.54}}{-9.844} [/tex]
t = -9.16 s or t = 12.75 s
According to the parabola that is the path of the falling object, the object is at zero height at t = -9.16 s and t = 12.75 s. Since the object starts moving at time t = 0, the domain has to be limited to t = 0 till t = 12.75 s.
Domain: [0, 12.75]
The range is the height of the object. Maximum height occurs at the time that is the midpoint of the two times the parabola shows h(t) = 0.
tmax = (-9.16 + 12.75)/2 = 1.797
Now we use the time of maximum height to find the maximum height.
h(t) = -4.922t^2 + 17.69t + 575
h(tmax) = -4.922(1.797)^2 + 17.69(1.797) + 575
h(tmax) = 590.9
The maximum height is 590.9 m.
Range: [0, 590.9]
