To find the quadratic function that models the given data, we need to determine a function of the form:
\[ y = ax^2 + bx + c \]
where \( a \), \( b \), and \( c \) are constants that we need to find.
We can use the data in the table to set up a system of equations. We'll pick three points from the data, specifically \((-2, 18)\), \((0, 0)\), and \((2, 3)\).
Using the points:
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For \( x = -2 \), \( y = 18 \): \[ 18 = a(-2)^2 + b(-2) + c \] \[ 18 = 4a - 2b + c \quad (1) \]
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For \( x = 0 \), \( y = 0 \): \[ 0 = a(0)^2 + b(0) + c \] \[ 0 = c \quad (2) \]
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For \( x = 2 \), \( y = 3 \): \[ 3 = a(2)^2 + b(2) + c \] \[ 3 = 4a + 2b + c \quad (3) \]
Using equation \( (2) \), we substitute \( c = 0 \) into equations \( (1) \) and \( (3) \):
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Substituting \( c = 0 \) into equation \( (1) \): \[ 18 = 4a - 2b \quad (4) \]
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Substituting \( c = 0 \) into equation \( (3) \): \[ 3 = 4a + 2b \quad (5) \]
Now we can solve equations \( (4) \) and \( (5) \):
From equation \( (4) \): \[ 4a - 2b = 18 \implies 2a - b = 9 \quad (6) \]
From equation \( (5) \): \[ 4a + 2b = 3 \implies 2a + b = \frac{3}{2} \quad (7) \]
Now, we can add equations \( (6) \) and \( (7) \): \[ (2a - b) + (2a + b) = 9 + \frac{3}{2} \] \[ 4a = 9 + 1.5 = 10.5 \implies a = \frac{10.5}{4} = 2.625 \]
Now substituting \( a \) back into equation \( (6) \) to find \( b \): \[ 2(2.625) - b = 9 \] \[ 5.25 - b = 9 \implies b = 5.25 - 9 = -3.75 \]
Now we have:
- \( a \approx 2.625 \)
- \( b \approx -3.75 \)
- \( c = 0 \)
Finally, the quadratic function is approximately: \[ y = 2.625x^2 - 3.75x \]
Now rounding to the nearest whole numbers: Typically, for many practical purposes, we represent the quadratic function as:
\[ y = \frac{21}{8}x^2 - \frac{15}{4}x \quad \text{(exact form)} \]
The final answer in decimal approximation is: \[ \text{The equation of the quadratic function} \quad y = 2.625 x^2 - 3.75 x \]
So, the full answer in function form would be:
\[ y = 2.625x^2 - 3.75x + 0 \]
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