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To determine the value of [tex]\( b \)[/tex] in the standard quadratic equation [tex]\( f(x) = ax^2 + bx + c \)[/tex], we need to use the given points to set up a system of equations and solve for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
Given the points [tex]\((x, f(x))\)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -4 & 7 \\ \hline -3 & 6 \\ \hline -2 & 7 \\ \hline -1 & 10 \\ \hline 0 & 15 \\ \hline 1 & 22 \\ \hline 2 & 31 \\ \hline \end{array} \][/tex]
We form the system of equations based on substituting each [tex]\( x \)[/tex] into the quadratic equation and setting it equal to [tex]\( f(x) \)[/tex].
1. For [tex]\( x = -4 \)[/tex]:
[tex]\[ 16a - 4b + c = 7 \][/tex]
2. For [tex]\( x = -3 \)[/tex]:
[tex]\[ 9a - 3b + c = 6 \][/tex]
3. For [tex]\( x = -2 \)[/tex]:
[tex]\[ 4a - 2b + c = 7 \][/tex]
4. For [tex]\( x = -1 \)[/tex]:
[tex]\[ a - b + c = 10 \][/tex]
5. For [tex]\( x = 0 \)[/tex]:
[tex]\[ c = 15 \][/tex]
6. For [tex]\( x = 1 \)[/tex]:
[tex]\[ a + b + c = 22 \][/tex]
7. For [tex]\( x = 2 \)[/tex]:
[tex]\[ 4a + 2b + c = 31 \][/tex]
Using the equation [tex]\( x = 0 \)[/tex]:
[tex]\[ c = 15 \][/tex]
We substitute [tex]\( c = 15 \)[/tex] into the remaining equations:
1. [tex]\( 16a - 4b + 15 = 7 \)[/tex]:
[tex]\[ 16a - 4b = -8 \quad \Rightarrow \quad 4a - b = -2 \quad \text{(Equation 1)} \][/tex]
2. [tex]\( 9a - 3b + 15 = 6 \)[/tex]:
[tex]\[ 9a - 3b = -9 \quad \Rightarrow \quad 3a - b = -3 \quad \text{(Equation 2)} \][/tex]
3. [tex]\( 4a - 2b + 15 = 7 \)[/tex]:
[tex]\[ 4a - 2b = -8 \quad \Rightarrow \quad 2a - b = -4 \quad \text{(Equation 3)} \][/tex]
4. [tex]\( a - b + 15 = 10 \)[/tex]:
[tex]\[ a - b = -5 \quad \text{(Equation 4)} \][/tex]
5. [tex]\( a + b + 15 = 22 \)[/tex]:
[tex]\[ a + b = 7 \quad \text{(Equation 5)} \][/tex]
6. [tex]\( 4a + 2b + 15 = 31 \)[/tex]:
[tex]\[ 4a + 2b = 16 \quad \Rightarrow \quad 2a + b = 8 \quad \text{(Equation 6)} \][/tex]
Now we solve these equations:
From Equations 4 and 5, we have:
[tex]\[ a - b = -5 \][/tex]
[tex]\[ a + b = 7 \][/tex]
We add these two equations:
[tex]\[ (a - b) + (a + b) = -5 + 7 \][/tex]
[tex]\[ 2a = 2 \][/tex]
[tex]\[ a = 1 \][/tex]
Next, we substitute [tex]\( a = 1 \)[/tex] into [tex]\( a + b = 7 \)[/tex]:
[tex]\[ 1 + b = 7 \][/tex]
[tex]\[ b = 6 \][/tex]
Thus, the value of [tex]\( b \)[/tex] is:
[tex]\[ \boxed{6} \][/tex]
Given the points [tex]\((x, f(x))\)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -4 & 7 \\ \hline -3 & 6 \\ \hline -2 & 7 \\ \hline -1 & 10 \\ \hline 0 & 15 \\ \hline 1 & 22 \\ \hline 2 & 31 \\ \hline \end{array} \][/tex]
We form the system of equations based on substituting each [tex]\( x \)[/tex] into the quadratic equation and setting it equal to [tex]\( f(x) \)[/tex].
1. For [tex]\( x = -4 \)[/tex]:
[tex]\[ 16a - 4b + c = 7 \][/tex]
2. For [tex]\( x = -3 \)[/tex]:
[tex]\[ 9a - 3b + c = 6 \][/tex]
3. For [tex]\( x = -2 \)[/tex]:
[tex]\[ 4a - 2b + c = 7 \][/tex]
4. For [tex]\( x = -1 \)[/tex]:
[tex]\[ a - b + c = 10 \][/tex]
5. For [tex]\( x = 0 \)[/tex]:
[tex]\[ c = 15 \][/tex]
6. For [tex]\( x = 1 \)[/tex]:
[tex]\[ a + b + c = 22 \][/tex]
7. For [tex]\( x = 2 \)[/tex]:
[tex]\[ 4a + 2b + c = 31 \][/tex]
Using the equation [tex]\( x = 0 \)[/tex]:
[tex]\[ c = 15 \][/tex]
We substitute [tex]\( c = 15 \)[/tex] into the remaining equations:
1. [tex]\( 16a - 4b + 15 = 7 \)[/tex]:
[tex]\[ 16a - 4b = -8 \quad \Rightarrow \quad 4a - b = -2 \quad \text{(Equation 1)} \][/tex]
2. [tex]\( 9a - 3b + 15 = 6 \)[/tex]:
[tex]\[ 9a - 3b = -9 \quad \Rightarrow \quad 3a - b = -3 \quad \text{(Equation 2)} \][/tex]
3. [tex]\( 4a - 2b + 15 = 7 \)[/tex]:
[tex]\[ 4a - 2b = -8 \quad \Rightarrow \quad 2a - b = -4 \quad \text{(Equation 3)} \][/tex]
4. [tex]\( a - b + 15 = 10 \)[/tex]:
[tex]\[ a - b = -5 \quad \text{(Equation 4)} \][/tex]
5. [tex]\( a + b + 15 = 22 \)[/tex]:
[tex]\[ a + b = 7 \quad \text{(Equation 5)} \][/tex]
6. [tex]\( 4a + 2b + 15 = 31 \)[/tex]:
[tex]\[ 4a + 2b = 16 \quad \Rightarrow \quad 2a + b = 8 \quad \text{(Equation 6)} \][/tex]
Now we solve these equations:
From Equations 4 and 5, we have:
[tex]\[ a - b = -5 \][/tex]
[tex]\[ a + b = 7 \][/tex]
We add these two equations:
[tex]\[ (a - b) + (a + b) = -5 + 7 \][/tex]
[tex]\[ 2a = 2 \][/tex]
[tex]\[ a = 1 \][/tex]
Next, we substitute [tex]\( a = 1 \)[/tex] into [tex]\( a + b = 7 \)[/tex]:
[tex]\[ 1 + b = 7 \][/tex]
[tex]\[ b = 6 \][/tex]
Thus, the value of [tex]\( b \)[/tex] is:
[tex]\[ \boxed{6} \][/tex]
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