Which one of the following is the second degree polynomial function f(x) where f(0) = 5. f(-1)= 10 and f(1) = 6?

Options

A. `5x^(2)-2x+5=0`
B. `3x^(2)-2x-5=0`
C. `3x^(2)-2x+5=0`
D. `3x^(2)-10x+5=0`

Answer

To solve the problem, we need to find a second degree polynomial function \( f(x) = ax^2 + bx + c \) that satisfies the following conditions:

1. \( f(0) = 5 \)
2. \( f(-1) = 10 \)
3. \( f(1) = 6 \)

### Step 1: Use the condition \( f(0) = 5 \)

Substituting \( x = 0 \) into the polynomial:

\[ f(0) = a(0)^2 + b(0) + c = c \]

Since \( f(0) = 5 \), we have:

\[ c = 5 \]

### Step 2: Use the condition \( f(-1) = 10 \)

Now substituting \( x = -1 \):

\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]

We already found \( c = 5 \), so:

\[ f(-1) = a - b + 5 \]

Setting this equal to 10 gives us:

\[ a - b + 5 = 10 \]
\[ a - b = 5 \quad \text{(Equation 1)} \]

### Step 3: Use the condition \( f(1) = 6 \)

Now substituting \( x = 1 \):

\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]

Again, substituting \( c = 5 \):

\[ f(1) = a + b + 5 \]

Setting this equal to 6 gives us:

\[ a + b + 5 = 6 \]
\[ a + b = 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations

Now we have a system of two equations:

1. \( a - b = 5 \) (Equation 1)
2. \( a + b = 1 \) (Equation 2)

We can solve these equations by adding them together:

\[ (a - b) + (a + b) = 5 + 1 \]
\[ 2a = 6 \]
\[ a = 3 \]

Now substitute \( a = 3 \) back into Equation 2:

\[ 3 + b = 1 \]
\[ b = 1 - 3 = -2 \]

### Step 5: Write the polynomial function

Now we have:

- \( a = 3 \)
- \( b = -2 \)
- \( c = 5 \)

Thus, the polynomial function is:

\[ f(x) = 3x^2 - 2x + 5 \]

### Verification:

1. \( f(0) = 5 \) (correct)
2. \( f(-1) = 3(-1)^2 - 2(-1) + 5 = 3 + 2 + 5 = 10 \) (correct)
3. \( f(1) = 3(1)^2 - 2(1) + 5 = 3 - 2 + 5 = 6 \) (correct)

### Conclusion

The second degree polynomial function is:

\[ f(x) = 3x^2 - 2x + 5 \]

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