Question

allie is playing basketball. she takes a shot 25 ft away from the basket. when the ball is 3 ft away from her, it is at a height of 10 ft above the floor. the ball reaches its highest height of 20 ft above the floor when it is 13 feet away from her.
a. find the value of a
(simplify your answer)
the value of a is $-\frac{1}{10}$.
b. if the hoop is 10 ft high, how close would allie have to be in order to make the basket?
b. allie would have to be \\(\square\\) feet from the basket
(type a whole number)

Explanation:

We assume the path of the basketball is a quadratic function, where \(x\) is the horizontal distance from Allie, and \(y\) is the height of the ball. The vertex form of a quadratic is \(y = a(x - h)^2 + k\), where \((h,k)\) is the vertex (highest point).

Step1: Identify vertex coordinates

The maximum height is 20 ft, when the ball is 13 ft from Allie, so \((h,k)=(13,20)\). Substitute into vertex form:
\(y = a(x - 13)^2 + 20\)

Step2: Use known point to find \(a\)

When \(x=3\), \(y=10\). Substitute these values:
\(10 = a(3 - 13)^2 + 20\)
Simplify inside the parentheses:
\(10 = a(-10)^2 + 20\)
\(10 = 100a + 20\)
Solve for \(a\):
\(100a = 10 - 20 = -10\)
\(a = \frac{-10}{100} = -\frac{1}{10}\)

Step3: Set up equation for part b

The basket is 25 ft from Allie horizontally, so we need to find \(x\) when \(y=10\) (hoop height) and \(x<13\) (closer to Allie than the vertex). Use the quadratic equation:
\(10 = -\frac{1}{10}(x - 13)^2 + 20\)

Step4: Solve for \(x\)

Rearrange the equation:
\(-\frac{1}{10}(x - 13)^2 = 10 - 20 = -10\)
Multiply both sides by -10:
\((x - 13)^2 = 100\)
Take square roots:
\(x - 13 = \pm 10\)
We take the negative root for the closer distance:
\(x = 13 - 10 = 3\)
Now find distance from the basket: The basket is 25 ft from Allie, so distance from basket is \(25 - 3 = 22\) ft.

Answer:

a. \(-\frac{1}{10}\)
b. 22