To model the path of the golf ball using a quadratic equation, we can start with the general form of a quadratic equation:
\[ h(x) = ax^2 + bx + c \]
where \( h(x) \) is the height in yards and \( x \) is the distance in yards.
From the table, we have the following points:
- (0, 0)
- (56, 26)
- (112, 34)
- (168, 26)
- (224, 0)
Since we have a symmetric path (the height increases to a maximum at \( x = 112 \) and then decreases), we can use the vertex form of a quadratic equation, which is given by:
\[ h(x) = a(x - h)^2 + k \]
where \( (h, k) \) is the vertex of the parabola.
Here, the vertex is \( (112, 34) \), so we can rewrite the equation as:
\[ h(x) = a(x - 112)^2 + 34 \]
Next, we can use another point from the table to find the value of \( a \). Let's use the point (56, 26).
Using this point in the equation, we have:
\[ 26 = a(56 - 112)^2 + 34 \]
Calculating \( 56 - 112 \):
\[ 56 - 112 = -56 \quad \Rightarrow \quad (-56)^2 = 3136 \]
Now substituting back into the equation:
\[ 26 = a(3136) + 34 \]
Subtracting 34 from both sides gives:
\[ 26 - 34 = 3136a \quad \Rightarrow \quad -8 = 3136a \]
Now we can solve for \( a \):
\[ a = \frac{-8}{3136} = -\frac{1}{392} \]
Calculating this value:
\[ a \approx -0.002551 \]
Rounding to the nearest thousandth:
\[ a \approx -0.003 \]
Therefore, in the equation that models this situation, \( a \) is approximately:
a = -0.003