Answer :
To solve for [tex]\( f(8) \)[/tex] for the quadratic function [tex]\( f(x) \)[/tex] given the points [tex]\( f(2)=6 \)[/tex], [tex]\( f(5)=12 \)[/tex], and [tex]\( f(11)=30 \)[/tex], we can follow the steps below:
### Step 1: Polynomial Form Setup
We know that a quadratic function takes the form:
[tex]\[ f(x) = ax^2 + bx + c \][/tex]
### Step 2: Establish the System of Equations
Using the given points, we can set up a system of equations:
[tex]\[
\begin{cases}
4a + 2b + c = 6 & \text{(1)} \\
25a + 5b + c = 12 & \text{(2)} \\
121a + 11b + c = 30 & \text{(3)}
\end{cases}
\][/tex]
### Step 3: Solve the System of Equations
Let's solve these equations step-by-step.
1. Subtract Equation (1) from Equation (2) to eliminate [tex]\( c \)[/tex]:
[tex]\[
(25a + 5b + c) - (4a + 2b + c) = 12 - 6
\][/tex]
[tex]\[
21a + 3b = 6 \quad \Rightarrow \quad 7a + b = 2 \quad \text{(4)}
\][/tex]
2. Subtract Equation (2) from Equation (3) to eliminate [tex]\( c \)[/tex]:
[tex]\[
(121a + 11b + c) - (25a + 5b + c) = 30 - 12
\][/tex]
[tex]\[
96a + 6b = 18 \quad \Rightarrow \quad 16a + b = 3 \quad \text{(5)}
\][/tex]
3. Subtract Equation (5) from Equation (4):
[tex]\[
(16a + b) - (7a + b) = 3 - 2
\][/tex]
[tex]\[
9a = 1 \quad \Rightarrow \quad a = \frac{1}{9}
\][/tex]
4. Substitute [tex]\( a = \frac{1}{9} \)[/tex] back into Equation (4):
[tex]\[
7a + b = 2 \quad \Rightarrow \quad 7 \left(\frac{1}{9}\right) + b = 2
\][/tex]
[tex]\[
\frac{7}{9} + b = 2 \quad \Rightarrow \quad b = 2 - \frac{7}{9} = \frac{18}{9} - \frac{7}{9} = \frac{11}{9}
\][/tex]
5. Substitute [tex]\( a = \frac{1}{9} \)[/tex] and [tex]\( b = \frac{11}{9} \)[/tex] back into Equation (1) to find [tex]\( c \)[/tex]:
[tex]\[
4a + 2b + c = 6
\][/tex]
[tex]\[
4\left(\frac{1}{9}\right) + 2\left(\frac{11}{9}\right) + c = 6
\][/tex]
[tex]\[
\frac{4}{9} + \frac{22}{9} + c = 6
\][/tex]
[tex]\[
\frac{26}{9} + c = 6 \quad \Rightarrow \quad c = 6 - \frac{26}{9} = \frac{54}{9} - \frac{26}{9} = \frac{28}{9}
\][/tex]
So the coefficients of the quadratic function are:
[tex]\[
a = \frac{1}{9}, \quad b = \frac{11}{9}, \quad c = \frac{28}{9}
\][/tex]
### Step 4: Determine [tex]\( f(8) \)[/tex]
Now, we substitute [tex]\( x = 8 \)[/tex] into the quadratic function:
[tex]\[
f(8) = a(8^2) + b(8) + c
\][/tex]
[tex]\[
f(8) = \frac{1}{9}(64) + \frac{11}{9}(8) + \frac{28}{9}
\][/tex]
[tex]\[
f(8) = \frac{64}{9} + \frac{88}{9} + \frac{28}{9} = \frac{64 + 88 + 28}{9} = \frac{180}{9} = 20
\][/tex]
So, the value of [tex]\( f(8) \)[/tex] is:
[tex]\[ \boxed{20} \][/tex]
### Step 1: Polynomial Form Setup
We know that a quadratic function takes the form:
[tex]\[ f(x) = ax^2 + bx + c \][/tex]
### Step 2: Establish the System of Equations
Using the given points, we can set up a system of equations:
[tex]\[
\begin{cases}
4a + 2b + c = 6 & \text{(1)} \\
25a + 5b + c = 12 & \text{(2)} \\
121a + 11b + c = 30 & \text{(3)}
\end{cases}
\][/tex]
### Step 3: Solve the System of Equations
Let's solve these equations step-by-step.
1. Subtract Equation (1) from Equation (2) to eliminate [tex]\( c \)[/tex]:
[tex]\[
(25a + 5b + c) - (4a + 2b + c) = 12 - 6
\][/tex]
[tex]\[
21a + 3b = 6 \quad \Rightarrow \quad 7a + b = 2 \quad \text{(4)}
\][/tex]
2. Subtract Equation (2) from Equation (3) to eliminate [tex]\( c \)[/tex]:
[tex]\[
(121a + 11b + c) - (25a + 5b + c) = 30 - 12
\][/tex]
[tex]\[
96a + 6b = 18 \quad \Rightarrow \quad 16a + b = 3 \quad \text{(5)}
\][/tex]
3. Subtract Equation (5) from Equation (4):
[tex]\[
(16a + b) - (7a + b) = 3 - 2
\][/tex]
[tex]\[
9a = 1 \quad \Rightarrow \quad a = \frac{1}{9}
\][/tex]
4. Substitute [tex]\( a = \frac{1}{9} \)[/tex] back into Equation (4):
[tex]\[
7a + b = 2 \quad \Rightarrow \quad 7 \left(\frac{1}{9}\right) + b = 2
\][/tex]
[tex]\[
\frac{7}{9} + b = 2 \quad \Rightarrow \quad b = 2 - \frac{7}{9} = \frac{18}{9} - \frac{7}{9} = \frac{11}{9}
\][/tex]
5. Substitute [tex]\( a = \frac{1}{9} \)[/tex] and [tex]\( b = \frac{11}{9} \)[/tex] back into Equation (1) to find [tex]\( c \)[/tex]:
[tex]\[
4a + 2b + c = 6
\][/tex]
[tex]\[
4\left(\frac{1}{9}\right) + 2\left(\frac{11}{9}\right) + c = 6
\][/tex]
[tex]\[
\frac{4}{9} + \frac{22}{9} + c = 6
\][/tex]
[tex]\[
\frac{26}{9} + c = 6 \quad \Rightarrow \quad c = 6 - \frac{26}{9} = \frac{54}{9} - \frac{26}{9} = \frac{28}{9}
\][/tex]
So the coefficients of the quadratic function are:
[tex]\[
a = \frac{1}{9}, \quad b = \frac{11}{9}, \quad c = \frac{28}{9}
\][/tex]
### Step 4: Determine [tex]\( f(8) \)[/tex]
Now, we substitute [tex]\( x = 8 \)[/tex] into the quadratic function:
[tex]\[
f(8) = a(8^2) + b(8) + c
\][/tex]
[tex]\[
f(8) = \frac{1}{9}(64) + \frac{11}{9}(8) + \frac{28}{9}
\][/tex]
[tex]\[
f(8) = \frac{64}{9} + \frac{88}{9} + \frac{28}{9} = \frac{64 + 88 + 28}{9} = \frac{180}{9} = 20
\][/tex]
So, the value of [tex]\( f(8) \)[/tex] is:
[tex]\[ \boxed{20} \][/tex]
