To determine the value of \( b \) in the standard form of the quadratic function \( f(x) = ax^2 + bx + c \), we can use the properties of quadratic functions and the given values in the table.
First, let's observe the values of \( f(x) \):
- \( f(-10) = 24 \)
- \( f(-9) = 17 \)
- \( f(-8) = 12 \)
- \( f(-7) = 9 \)
- \( f(-6) = 8 \)
- \( f(-5) = 9 \)
- \( f(-4) = 12 \)
Next, we can notice that \( f(x) \) has a vertex since it is quadratic, and the values drop to a minimum and then rise again. This suggests that the function is symmetrical.
To find the vertex, we can look for the average of the x-values at which the function reaches its minimum. Observing the values, the minimum seems to occur around \( x = -6 \) or \( x = -7 \).
Next, we can compute the differences between successive \( f(x) \) values to analyze the behavior of the function:
- From \( x = -10 \) to \( x = -9 \): \( 24 - 17 = 7 \)
- From \( x = -9 \) to \( x = -8 \): \( 17 - 12 = 5 \)
- From \( x = -8 \) to \( x = -7 \): \( 12 - 9 = 3 \)
- From \( x = -7 \) to \( x = -6 \): \( 9 - 8 = 1 \)
- From \( x = -6 \) to \( x = -5 \): \( 8 - 9 = -1 \)
- From \( x = -5 \) to \( x = -4 \): \( 9 - 12 = -3 \)
The differences show a symmetrical pattern around the minimum. This indicates that the minimum value of \( f(x) \) is indeed at \( x = -6 \) or \( x = -7 \).
To find the coefficients \( a \), \( b \), and \( c \), we can select points to create a system of equations.
Let's choose three points, say \( (-8, 12) \), \( (-6, 8) \), and \( (-4, 12) \).
Substituting these points into the standard form \( f(x) = ax^2 + bx + c \) gives us three equations:
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For \( (-8, 12) \): \[ 12 = a(-8)^2 + b(-8) + c \implies 12 = 64a - 8b + c \]
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For \( (-6, 8) \): \[ 8 = a(-6)^2 + b(-6) + c \implies 8 = 36a - 6b + c \]
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For \( (-4, 12) \): \[ 12 = a(-4)^2 + b(-4) + c \implies 12 = 16a - 4b + c \]
We can solve this system of equations step-by-step.
To eliminate \( c \), we can subtract the second equation from the first, and the third from the second:
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\( (64a - 8b + c) - (36a - 6b + c) = 12 - 8 \) \[ 28a - 2b = 4 \implies 14a - b = 2 \quad \text{(Equation 1)} \]
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\( (36a - 6b + c) - (16a - 4b + c) = 8 - 12 \) \[ 20a - 2b = -4 \implies 10a - b = -2 \quad \text{(Equation 2)} \]
Now we can solve Equations 1 and 2: From Equation 1: \[ b = 14a - 2 \]
Substituting into Equation 2: \[ 10a - (14a - 2) = -2 \] \[ 10a - 14a + 2 = -2 \] \[ -4a + 2 = -2 \] \[ -4a = -4 \implies a = 1 \]
Now substitute \( a = 1 \) back into Equation 1: \[ b = 14(1) - 2 = 12 \]
So the value of \( b \) is \( \boxed{12} \).
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