Certainly! Let's solve the given problem step by step.
We are given the function [tex]\( f(x) = ax^2 + bx + c \)[/tex] where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are constants.
### Step 1: Determine the value of [tex]\( c \)[/tex].
We are given that [tex]\( f(0) = 5 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = a \cdot 0^2 + b \cdot 0 + c = c \][/tex]
Thus:
[tex]\[ c = 5 \][/tex]
### Step 2: Use the values of [tex]\( f(1) \)[/tex] and [tex]\( f(-1) \)[/tex] to set up a system of equations to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
We are given:
[tex]\[ f(1) = 6 \][/tex]
[tex]\[ f(1) = a \cdot 1^2 + b \cdot 1 + c \][/tex]
Substituting the value of [tex]\( c \)[/tex] we found earlier:
[tex]\[ 6 = a \cdot 1 + b \cdot 1 + 5 \][/tex]
[tex]\[ 6 = a + b + 5 \][/tex]
Subtracting 5 from both sides:
[tex]\[ a + b = 1 \quad \text{(Equation 1)} \][/tex]
We are also given:
[tex]\[ f(-1) = 8 \][/tex]
[tex]\[ f(-1) = a \cdot (-1)^2 + b \cdot (-1) + c \][/tex]
Substituting the value of [tex]\( c \)[/tex]:
[tex]\[ 8 = a \cdot 1 - b \cdot 1 + 5 \][/tex]
[tex]\[ 8 = a - b + 5 \][/tex]
Subtracting 5 from both sides:
[tex]\[ a - b = 3 \quad \text{(Equation 2)} \][/tex]
### Step 3: Solve the system of equations.
We have the two equations:
1. [tex]\( a + b = 1 \)[/tex]
2. [tex]\( a - b = 3 \)[/tex]
Add these two equations:
[tex]\[ (a + b) + (a - b) = 1 + 3 \][/tex]
[tex]\[ 2a = 4 \][/tex]
Divide by 2:
[tex]\[ a = 2 \][/tex]
Substitute [tex]\( a = 2 \)[/tex] back into Equation 1:
[tex]\[ 2 + b = 1 \][/tex]
Subtract 2 from both sides:
[tex]\[ b = -1 \][/tex]
### Conclusion
The constants in the quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] are:
- [tex]\( c = 5 \)[/tex]
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -1 \)[/tex]
Thus, the values of the constants are:
[tex]\[ c = 5, \quad a = 2, \quad b = -1 \][/tex]
