Question

For quadratic function f, the solutions to the equation f(x) = 0 are x = 8 and x = –4. Which function could represent f ?


{silence}Option A.{silence}
f(x) = x2 + 4x – 32


{silence}Option Be.{silence}
f(x) = x2 + 12x + 32


{silence}Option C.{silence}
f(x) = x2 – 4x – 32


{silence}Option D.{silence}
f(x) = x2 – 12x + 32

Answer 1

To find a quadratic function \( f(x) \) with the given roots \( x = 8 \) and \( x = -4 \), we can use the fact that if a quadratic function has roots \( r_1 \) and \( r_2 \), the function can be expressed in factored form as:

\[
f(x) = a(x - r_1)(x - r_2)
\]

Substituting the roots \( r_1 = 8 \) and \( r_2 = -4 \):

\[
f(x) = a(x - 8)(x + 4)
\]

Next, we can expand this expression:

\[
f(x) = a[(x - 8)(x + 4)]
\]

\[
= a[x^2 + 4x - 8x - 32]
\]

\[
= a[x^2 - 4x - 32]
\]

Assuming \( a = 1 \) for simplicity, we get:

\[
f(x) = x^2 - 4x - 32
\]

Now, let's compare this with the given options:

- **Option A:** \( f(x) = x^2 + 4x - 32 \) (not a match)
- **Option B:** \( f(x) = x^2 + 12x + 32 \) (not a match)
- **Option C:** \( f(x) = x^2 - 4x - 32 \) (this matches)
- **Option D:** \( f(x) = x^2 - 12x + 32 \) (not a match)

Therefore, the quadratic function that could represent \( f \) is:

**Option C: \( f(x) = x^2 - 4x - 32 \)**.