To find the equation of the quadratic that represents the height of the golf ball as a function of distance from Lorraine, we can use the vertex form of a quadratic equation:
\[ f(x) = a(x - h)^2 + k \]
Here, \((h, k)\) is the vertex of the parabola. From the information given:
- The vertex is \((30, 30)\) because at the highest point (the vertex), the ball is 30 feet high and is 30 feet away from Lorraine.
- The ball lands 60 feet away, indicating that the \(x\)-intercepts are at \(x = 0\) (where it starts) and \(x = 60\) (where it lands).
To find the value of \(a\), we can use one of the intercepts. We know \(f(0) = 0\) and \(f(60) = 0\). We can use the vertex to find \(a\) by plugging in the height of the vertex into the quadratic equation using one of the x-intercepts.
Using \((0, 0)\):
\[ 0 = a(0 - 30)^2 + 30 \] \[ 0 = 900a + 30 \] \[ 900a = -30 \] \[ a = -\frac{1}{30} \]
Now we can substitute \(a\), \(h\), and \(k\) into the vertex form:
\[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
This is the equation of the scenario.
Next, we need to find the height of the ball when it is 15 feet away from Lorraine. We can use \(x = 15\):
\[ f(15) = -\frac{1}{30}(15 - 30)^2 + 30 \] \[ f(15) = -\frac{1}{30}(-15)^2 + 30 \] \[ f(15) = -\frac{1}{30}(225) + 30 \] \[ f(15) = -\frac{225}{30} + 30 \] \[ f(15) = -7.5 + 30 \] \[ f(15) = 22.5 \]
So, the equation for this scenario is:
\[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
The height of the ball when it is 15 feet away from Lorraine is:
22.5 feet.