Answer:
Part 1) The maximum height of the softball is 18 feet
Part 2) The range for H(t) is the interval [0,18]
[tex]0\leq H(t)\leq 18[/tex]
Part 3) The domain is the interval [0,2.06]
[tex]0\leq t\leq 2.06[/tex]
Step-by-step explanation:
step 1
Find the maximum height of the softball
we have
[tex]H(t)=-16t^2+32t+2[/tex]
This is a vertical parabola open downward
The vertex represent a maximum
step 1
Find the maximum height of the ball
we know that
The maximum height of the ball correspond to the y-coordinate of the vertex
Convert the quadratic equation into vertex form
Factor -16
[tex]H(t)=-16(t^2-2t)+2[/tex]
Complete the square
[tex]H(t)=-16(t^2-2t+1)+2+16[/tex]
[tex]H(t)=-16(t^2-2t+1)+18[/tex]
Rewrite as perfect squares
[tex]H(t)=-16(t-1)^2+18[/tex]
The vertex is the point (1,18)
therefore
The maximum height of the softball is 18 feet
step 2
Find the range of the function
The range for H(t) is the interval [0,18]
[tex]0\leq H(t)\leq 18[/tex]
step 3
Find the domain
To find out the domain we need to determine the x-intercepts
For H(t)=0
[tex]0=-16(t-1)^2+18[/tex]
[tex]16(t-1)^2=18[/tex]
[tex](t-1)^2=\frac{18}{16}[/tex]
[tex]t-1=\pm\frac{3\sqrt{2}}{4}[/tex]
[tex]t=\pm\frac{3\sqrt{2}}{4}+1[/tex]
The solutions are
[tex]t=\frac{3\sqrt{2}}{4}+1=2.06\ sec[/tex]
[tex]t=-\frac{3\sqrt{2}}{4}+1=-0.06\ sec[/tex] ---> is not solution because the time cannot be negative
so
The domain is the interval [0,2.06]
[tex]0\leq t\leq 2.06[/tex]
