To find the quadratic function that best fits the given data points, we can use the general form of a quadratic equation:
\[ y = ax^2 + bx + c \]
We have the following data points:
- (62, 104)
- (64, 34)
- (67, 4)
- (69, 29)
- (71, 98)
We can set up a system of equations based on these data points. Since we have three unknowns \(a\), \(b\), and \(c\), we can choose three points to form equations. However, since we have five data points, we might typically use polynomial regression or another method to find the best-fitting coefficients, but for simplicity, let's proceed using just three points from the dataset.
Let's select the points (62, 104), (64, 34), and (67, 4) for our system of equations:
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From the point \( (62, 104) \): \[ 104 = a(62^2) + b(62) + c \] This simplifies to: \[ 104 = 3844a + 62b + c \tag{1} \]
-
From the point \( (64, 34) \): \[ 34 = a(64^2) + b(64) + c \] This simplifies to: \[ 34 = 4096a + 64b + c \tag{2} \]
-
From the point \( (67, 4) \): \[ 4 = a(67^2) + b(67) + c \] This simplifies to: \[ 4 = 4489a + 67b + c \tag{3} \]
Now we can subtract equation (1) from (2) and (2) from (3) to eliminate \(c\):
From (2) - (1): \[ 34 - 104 = (4096 - 3844)a + (64 - 62)b \] \[ -70 = 252a + 2b \] \[ -35 = 126a + b \tag{4} \]
From (3) - (2): \[ 4 - 34 = (4489 - 4096)a + (67 - 64)b \] \[ -30 = 393a + 3b \] \[ -10 = 131a + b \tag{5} \]
Next, we subtract (4) from (5): \[ (-10) - (-35) = (131a - 126a) + (b - b) \] \[ 25 = 5a \] \[ a = 5 \]
Now we can use the value of \(a\) to find \(b\). Substitute \(a\) into equation (4): \[ -35 = 126(5) + b \] \[ -35 = 630 + b \] \[ b = -35 - 630 \] \[ b = -665 \]
Now, rounding \(b\) to the nearest whole number:
The value of \(b\) is: \[ b = -665 \]
So, in the quadratic function that represents the dataset, \(b\) is \(-665\) rounded to the nearest whole number.