Chapter 4.3, Problem 16PPSE

Solution

To Find: If each ball pass through the hoop if the centre of the hoop is located at $\left(12,10\right)$

For the given parabolic function each pass through the hoop.

For the given table each pass will not pass through the hoop.

Given:

The equation $y=-0.125x^2+1.84x+6$

Centre of the hoop: $\left(12,10\right)$

Another basket ball model:

$x$	$y$
$2$	$\begin{array}{r}\hfill 10\end{array}$
$4$	$12$
$10$	$12$

Explanation:

Find whether the ball pass through the hoop for the given model:

  $y=-0.125x^2+1.84x+6$

Substitute the given centre of the hoop $x=12,y=10$ in the above model:

  $\begin{array}{l}10=-0.125{\left(12\right)}^2+1.84\left(12\right)+6\\ 10=-\text{18+22}\text{.08+6}\\ 10\approx 10.08\end{array}$

Thus, the ball pass through the hoop for the given function:

Find the ball which pass through for another given table:

$x$	$y$
$2$	$\begin{array}{r}\hfill 10\end{array}$
$4$	$12$
$10$	$12$

Find the parabolic equation:

Use the standard form of a quadratic equation $y=a{x}^2+bx+c$ as the starting point for finding the equation through the three points.

  $y=a{x}^2+bx+c$

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three-equation system.

  $\begin{array}{l}10=a{\left(2\right)}^2+b\left(2\right)+c\to \left(1\right)\\ 12=a{\left(4\right)}^2+b\left(4\right)+c\to \left(2\right)\\ 12=a{\left(10\right)}^2+b\left(10\right)+c\to \left(3\right)\end{array}$

Solve for $c$ in equation (1):

  $c=10-4a-2b$

Replace all occurrences of $c$ with $10-4a-2b$ in each equation:

  $\begin{array}{l}12=96a+8b+10\hfill \\ 12=12a+2b+10\hfill \\ c=10-4a-2b\hfill \end{array}$

Solve for $a$ in $12=96a+8b+10$ :

  $\begin{array}{c}a=-\frac{b}{12}+\frac{1}{48}\\ 12=12a+2b+10\\ c=10-4a-2b\end{array}$

Replace all occurrences of $a$ with $-\frac{b}{12}+\frac{1}{48}$ in each equation:

  $\begin{array}{c}c=-\frac{5b}{3}+\frac{119}{12}\\ 12=b+\frac{41}{4}\\ a=-\frac{b}{12}+\frac{1}{48}\end{array}$

Solve for $b$ in $12=b+\frac{41}{4}$

  $\begin{array}{c}b=\frac{7}{4}\\ c=-\frac{5b}{3}+\frac{119}{12}\\ a=-\frac{b}{12}+\frac{1}{48}\end{array}$

Replace all occurrences of $b$ with $\frac{7}{4}$ in each equation:

  $a=-\frac{1}{8},c=7,b=\frac{7}{4}$

Substitute the actual values into the formula for a quadratic equation:

  $y=-\frac{x^2}{8}+\frac{7x}{4}+7$

From the table, substitute $x=2$ and $y=10$ in the equation $y=-\frac{x^2}{8}+\frac{7x}{4}+7$ :

  $\begin{array}{c}10=-\frac{2^2}{8}+\frac{7\left(2\right)}{4}+7\\ 10=-\frac{1}{2}+\frac{7}{2}+7\\ 10=10\end{array}$

Substitute $x=4$ and $y=10$ in the equation $y=-\frac{x^2}{8}+\frac{7x}{4}+7$ :

  $\begin{array}{l}10=-0.125{\left(4\right)}^2+1.84\left(4\right)+6\\ 10=-\text{2+7+7}\\ 10\ne 12\end{array}$

Substitute $x=10$ and $y=10$ in the equation $y=-\frac{x^2}{8}+\frac{7x}{4}+7$ :

  $\begin{array}{l}10=-\frac{{10}^2}{8}+\frac{7\left(10\right)}{4}+7\\ 10=-\frac{25}{2}+\frac{35}{2}+7\\ 10\ne 12\end{array}$

Thus, for the given table each pass will not pass through the hoop.