Chapter 4.4, Problem 97PPSE

Solution

a.

To Find: The quadratic model for the ball's height as a function of time.

The quadratic model is $y=-16x^2+22x+3$ .

Given: $$

The height of the ball at certain times appears in the table below.

Time $\left(s\right)$	$0.5$	$0.75$	$1$	$1.25$
Height $\left(ft\right)$	$10$	$10.5$	$9$	$5.5$

Explanation:

Given points are $\left(0.5,10\right),\left(0.75,10.5\right),\left(1,9\right)$ and $\left(1.25,5.5\right)$ :

Substitute the first point $\left(0.5,10\right)\to \left(x,y\right)$

  $\begin{array}{c}y=a{x}^2+bx+c\\ 10=a{\left(0.5\right)}^2+b\left(0.5\right)+c\\ 10=0.25a+0.5b+c\to \left(1\right)\end{array}$

Substitute the second point $\left(0.75,10.5\right)\to \left(x,y\right)$

  $\begin{array}{c}y=a{x}^2+bx+c\\ 10.5=a{\left(0.75\right)}^2+b\left(0.75\right)+c\\ 10.5=0.5625a+0.75b+c\to \left(2\right)\end{array}$

Substitute the third point $\left(1,9\right)\to \left(x,y\right),$

  $\begin{array}{c}y=a{x}^2+bx+c\\ 9=a{\left(1\right)}^2+b\left(1\right)+c\\ 9=a+b+c\to \left(3\right)\end{array}$

Substitute the fourth point $\left(1.25,5.5\right)\to \left(x,y\right)$

  $\begin{array}{c}y=a{x}^2+bx+c\\ 5.5=a{\left(1.25\right)}^2+b\left(1.25\right)+c\\ 5.5=1.5625a+1.25b+c\to \left(4\right)\end{array}$

Multiply equation $1\times \left(-1\right)$ :

  $-10=-025a-0.5b-c\to \left(1.1\right)$

Elimination between equation $\left(1.1\right)$ and $\left(2\right)$ :

  $\begin{array}{l}-10=-0.25a-0.5b-c\\ \underset{¯}{+10.5=0.5625a+0.75b+c}\\ 0.5=0.3125a+0.25b\to \left(5\right)\end{array}$

Multiply equation $3\times \left(-1\right)$ :

  $-9=-a-b-c$

Elimination between equation $\left(3.1\right)$ and $\left(4\right)$ :

  $\begin{array}{l}\underset{¯}{\begin{array}{cc}-9=& -a-b-c\\ 5.5=& 1.5625a+1.25b+c\end{array}}\\ -3.5=0.5625a+0.25b\to \left(6\right)\end{array}$

Multiply equation $5\times \left(-1\right)$ :

  $-0.5=-0.3125a-0.25b\to \left(5.1\right)$

Elimination between equation $\left(5.1\right)$ and $\left(6\right)$ :

  $\begin{array}{l}-0.5=-0.3125a-0.25b\\ \underset{¯}{-3.5=0.5625a+0.25b}\\ -4=0.25a\to \left(6\right)\end{array}$

Solve for $a$ :

  $\begin{array}{c}-4=0.25a\\ -16=a\end{array}$

Substitute $a=-16$ in equation $5$ , to find for the value of $b$ :

  $\begin{array}{l}-0.5=-0.3125a-0.25b\hfill \\ -0.5=-0.3125\left(-16\right)-0.25b\hfill \\ -0.5=5-0.25b\hfill \\ -5.5=-0.25b\hfill \\ 22=b\hfill \end{array}$

Now, substitute $a=-16$ and $b=22$ in equation $3$ to find $c$ :

  $\begin{array}{c}9=a+b+c\\ 9=-16+22+c\\ 9=6+c\\ 3=c\end{array}$

Substitute $a=-16,b=22$ and $c=3,$ in $y=a{x}^2+bx+c$ :

  $y=-16x^2+22x+3$

Thus, the quadratic model is $y=-16x^2+22x+3$ .

b.

To Write: The quadratic function in factored form.

The factored form is $\left(2x-3\right)\left(8x+1\right)$

Given: $$

The height of the ball at certain times appears in the table below.

Time $\left(s\right)$	$0.5$	$0.75$	$1$	$1.25$
Height $\left(ft\right)$	$10$	$10.5$	$9$	$5.5$

Explanation:

From part $\left(a\right)$

  $y=-16x^2+22x+3$

Factor the expression $-16x^2+22x+3=0$

  $\begin{array}{c}-16x^2+22x+3=0\\ 16x^2-22x-3=0\\ 16x^2+2x-24x-3=0\\ 2x\left(8x+1\right)-3\left(8x+1\right)=0\\ \left(2x-3\right)\left(8x+1\right)=0\end{array}$

Therefore, the factored form is $\left(2x-3\right)\left(8x+1\right)$