Question

for quadratic function f, the solutions to the equation f(x) = 0 are x = 8 and x = -4. which function could represent f

Answer 1

Given that the solutions to the equation \( f(x) = 0 \) are \( x = 8 \) and \( x = -4 \), we can express the quadratic function in its factored form as follows:

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

where \( a \) is a non-zero constant that determines the vertical stretch/compression and direction (upwards or downwards) of the parabola.

To find one specific function that could represent \( f \), we can choose \( a = 1 \):

\[
f(x) = 1(x - 8)(x + 4) = (x - 8)(x + 4)
\]

Now multiply the factors:

\[
f(x) = x^2 + 4x - 8x - 32 = x^2 - 4x - 32
\]

Thus, one possible quadratic function that represents \( f \) is:

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

You could also choose other values for \( a \), such as \( a = 2 \), \( a = -1 \), etc., leading to different quadratic functions, but the roots will remain the same. For example:

If \( a = 2 \):
\[
f(x) = 2(x - 8)(x + 4) = 2x^2 - 8x - 64
\]

If \( a = -1 \):
\[
f(x) = -1(x - 8)(x + 4) = -(x^2 - 4x - 32) = -x^2 + 4x + 32
\]

All of these functions would have the same roots \( x = 8 \) and \( x = -4 \).

Answer 2

Answer using these choices: a. f(x) = x^2 + 4x - 32
b. f(x) = x^2 + 12x + 32
c. f(x) = x^2 - 4x - 32
d. f(x) = x^2 - 12x + 32