Chapter 8.3, Problem 23Q

To determine

To find: the maximum height of the softball and also describe its domain and range.

Answer to Problem 23Q

Maximum height = 18 feet

Domain: $\left[0,2.061\right]$

Range: $\left[0,18\right]$

Explanation of Solution

Given information:

The height (in feet) of a softball t seconds after it is pitched is given by the function:

  $h\left(t\right)=-16t^2+32t+2$ ….. (1)

Concept Used:

Maximum and Minimum value:

The y -coordinate of the vertex of the graph of $f\left(x\right)=a{x}^2+bx+c$ is the maximum value if the function when $a<0$ or the minimum value of the function when $a>0$ .

Vertex $\left(x_0,y_0\right)$ of the function $f\left(x\right)=a{x}^2+bx+c$ I given by:

  $x_0=-\frac{b}{2a}$

And

  $y_0=f\left(x_0\right)$

Calculation:

Comparing the given function $h\left(t\right)=-16t^2+32t+2$ with $f\left(x\right)=a{x}^2+bx+c$ , we have that

  $a=-16$ , $b=32$ And $c=2$

First find the vertex $\left(x_0,y_0\right)$ of the given parabola:

  $\begin{array}{c}{x}_0=-\frac{b}{2a}\\ =-\frac{32}{2\left(-16\right)}\\ =1\end{array}$

  $\begin{array}{c}{y}_0=h\left(x_0\right)\\ =-16{\left(1\right)}^2+32\left(1\right)+2\\ =-16+32+2\\ =18\end{array}$

So, the vertex is $\left(1,18\right)$

Since, here $a=-16<0$

So, the maximum height of the softball is the y -coordinate of the vertex that is, maximum height is 18 feet.

The graph of the given function is:

  

Domain of the given function is the time from when the ball is pitched to the time it hits the ground.

By observing the graph it can be said that:

Domain: $\left[0,2.061\right]$

Range: $\left[0,18\right]$