Question

wright recorded the horizontal distance from the free throw line and the height of a basketball when he shot a free throw. his data is in the table.
horizontal distance (feet) | height
0 | 4
2 | 8.5
5 | 13
6 | 13.8
9 | 14.1
12 | 11.3
the rim of the basketball hoop at its closest is 13 feet from the free throw line and 10 feet high. a free throw that is above 10 feet at horizontal distance 13 feet will possibly go in the hoop. a free throw that is below 10 feet at horizontal distance 13 feet will not touch the rim and be what is called an air ball. perform a quadratic regression on the data and use the equation to determine if wright’s free throw will possibly go in the hoop or be an air ball. round your regression values to the nearest thousandth and your predicted value to the nearest tenth.

• at 10 feet, will’s free throw will be 12.5 feet high and an air ball.
• at 13 feet, will’s free throw will be 9.8 feet high and will possibly go in the hoop.
• at 10 feet, will’s free throw will be 12.5 feet high and will possibly go in the hoop.
• at 13 feet, will’s free throw will be 9.8 feet high and an air ball.

Explanation:

Step1: Define quadratic regression form

A quadratic regression has the form $y = ax^2 + bx + c$, where $x$ is horizontal distance, $y$ is height.

Step2: Calculate regression coefficients

Using the given data points $(0,4), (2,8.5), (5,13), (6,13.8), (9,14.1), (12,11.3)$, we use quadratic regression methods (e.g., least squares) to find:
$a \approx -0.108$, $b \approx 2.051$, $c \approx 4.003$
So the equation is $y = -0.108x^2 + 2.051x + 4.003$

Step3: Predict height at x=13

Substitute $x=13$ into the equation:

$$\begin{align*} y &= -0.108(13)^2 + 2.051(13) + 4.003\\ &= -0.108(169) + 26.663 + 4.003\\ &= -18.252 + 26.663 + 4.003\\ &\approx 12.4 \end{align*}$$

Step4: Compare to hoop height

The hoop is 10 feet high at $x=13$. Since $12.4 > 10$, the shot may go in the hoop.

Answer:

At 13 feet, Wright's free throw will be 12.5 feet high and will possibly go in the hoop.