Question

The function shown models the height (in feet) of a softball 1 seconds after it

Answer 1

Answer:

Part 1) The maximum height of the softball is 18 feet

Part 2) The range for H(t) is the interval [0,18]

$0\leq H(t)\leq 18$

Part 3) The domain is the interval [0,2.06]

$0\leq t\leq 2.06$

Step-by-step explanation:

step 1

Find the maximum height of the softball

we have

$H(t)=-16t^2+32t+2$

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

$H(t)=-16(t^2-2t)+2$

Complete the square

$H(t)=-16(t^2-2t+1)+2+16$

$H(t)=-16(t^2-2t+1)+18$

Rewrite as perfect squares

$H(t)=-16(t-1)^2+18$

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]

$0\leq H(t)\leq 18$

step 3

Find the domain

To find out the domain we need to determine the x-intercepts

For H(t)=0

$0=-16(t-1)^2+18$

$16(t-1)^2=18$

$(t-1)^2=\frac{18}{16}$

$t-1=\pm\frac{3\sqrt{2}}{4}$

$t=\pm\frac{3\sqrt{2}}{4}+1$

The solutions are

$t=\frac{3\sqrt{2}}{4}+1=2.06\ sec$

$t=-\frac{3\sqrt{2}}{4}+1=-0.06\ sec$ ---> is not solution because the time cannot be negative

so

The domain is the interval [0,2.06]

$0\leq t\leq 2.06$