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Follow steps similar to Example 1. This time the vertex is at \((7, 16)\), therefore \(p = 7\) and \(q = 16\).

Write an initial equation in vertex form, using \(p = 7\) and \(q = 16\).

\(\begin{array}{l}
 f\left( x \right) = a\left( {x - p} \right)^2 + q \\
 f\left( x \right) = a\left( {x - 7} \right)^2 + 16 \end{array}\)

Using the ball’s starting point, \((15, 7)\), substitute \(x = 15\) and \(f(x) = 7\) into the equation, and solve for \(a\).

\(\begin{align}
 f\left( x \right) &= a\left( {x - 4} \right)^2 + 12 \\
 7 &= a\left( {15 - 7} \right)^2 + 16 \\
  - 9 &= 64a \\
 0.140 \thinspace 625 &= a \end{align}\)

The equation of the function is

\(f(x) = -0.140 \thinspace 625(x - 7)^2 + 16\)

Then, solve for the \(y\)-intercept by substituting \(x = 0\) into the equation.

\(\begin{array}{l}
 f\left( x \right) = - 0.140 \thinspace 625\left( {x - 7} \right)^2 + 16 \\
 f\left( 0 \right) = - 0.140 \thinspace 625\left( {0 - 7} \right)^2 + 16 \\
 f\left( 0 \right) = 9.109 \thinspace 375  \end{array}\)

This is under \(10\) ft.

Next, solve for the value of \(y\), given \(x = 1.5\).

\(\begin{align}
 f\left( x \right) &= - 0.140 \thinspace 625\left( {x - 7} \right)^2 + 16 \\
 f\left( {1.5} \right) &= - 0.140 \thinspace 625\left( {1.5 - 7} \right)^2 + 16 \\
 f\left( {1.5} \right) &= 11.746... \end{align}\)

Because the ball crosses over the value of \(y = 10\) between \(x = 0\) and \(x = 1.5\). Huston has likely made a basket!