Question

part 1 of 2
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.
b. if the hoop is 10 ft high, how close would allie have to be in order to make the basket?
a. the value of a is
(simplify your answer.)

Explanation:

We assume the path of the basketball is a quadratic function \(y = a(x - h)^2 + k\), where \((h,k)\) is the vertex (highest point).

Step1: Identify vertex coordinates

The highest point is 20 ft high, and it occurs when the ball is 13 ft from Allie, so \(h=13\), \(k=20\). The function is \(y = a(x - 13)^2 + 20\).

Step2: Use given point to solve for a

When \(x=3\), \(y=10\). Substitute into the function:
\(10 = a(3 - 13)^2 + 20\)
\(10 = a(-10)^2 + 20\)
\(10 = 100a + 20\)

Step3: Solve equation for a

Subtract 20 from both sides:
\(10 - 20 = 100a\)
\(-10 = 100a\)
Divide by 100:
\(a = \frac{-10}{100} = -\frac{1}{10}\)

Step4: Set up equation for part b

The hoop is 10 ft high (\(y=10\)), and we need to find \(x\) (distance from Allie) when the ball goes into the hoop. Use the quadratic function:
\(10 = -\frac{1}{10}(x - 13)^2 + 20\)

Step5: Solve for \(x\)

Subtract 20 from both sides:
\(10 - 20 = -\frac{1}{10}(x - 13)^2\)
\(-10 = -\frac{1}{10}(x - 13)^2\)
Multiply both sides by -10:
\(100 = (x - 13)^2\)
Take square roots:
\(x - 13 = \pm 10\)
So \(x = 13 + 10 = 23\) or \(x = 13 - 10 = 3\). We discard \(x=3\) (this is the given point), so \(x=23\). Since the basket is 25 ft from Allie initially, the distance from Allie to the basket for the shot is \(25 - 23 = 2\) ft.

Answer:

a. \( -\frac{1}{10} \)
b. 2 feet