Certainly! Let's analyze each function one by one to determine which one meets the criteria of having [tex]$x$[/tex]-intercepts at [tex]$(0,0)$[/tex] and [tex]$(4,0)$[/tex]. The [tex]$x$[/tex]-intercepts of a function are the points where the function crosses the [tex]$x$[/tex]-axis, which occurs where [tex]$f(x) = 0$[/tex].
1. \( f(x) = x(x - 4) \)
For the function \( f(x) = x(x - 4) \), we set the function equal to zero to find the [tex]$x$[/tex]-intercepts:
[tex]\[
x(x - 4) = 0
\][/tex]
This equation holds when either \( x = 0 \) or \( x - 4 = 0 \). Solving \( x - 4 = 0 \), we get:
[tex]\[
x = 4
\][/tex]
So, the intercepts are at \( (0,0) \) and \( (4,0) \).
2. \( f(x) = x(x+4) \)
For the function \( f(x) = x(x + 4) \), we set the function equal to zero to find the [tex]$x$[/tex]-intercepts:
[tex]\[
x(x + 4) = 0
\][/tex]
This equation holds when either \( x = 0 \) or \( x + 4 = 0 \). Solving \( x + 4 = 0 \), we get:
[tex]\[
x = -4
\][/tex]
So, the intercepts are at \( (0,0) \) and \( (-4,0) \).
3. \( f(x) = (x - 4)(x - 4) \)
For the function \( f(x) = (x - 4)(x - 4) \), we set the function equal to zero to find the [tex]$x$[/tex]-intercepts:
[tex]\[
(x - 4)(x - 4) = 0
\][/tex]
This equation holds when \( x - 4 = 0 \). Solving \( x - 4 = 0 \), we get:
[tex]\[
x = 4
\][/tex]
This gives a repeated intercept at \( (4,0) \).
4. \( f(x) = (x + 4)(x + 4) \)
For the function \( f(x) = (x + 4)(x + 4) \), we set the function equal to zero to find the [tex]$x$[/tex]-intercepts:
[tex]\[
(x + 4)(x + 4) = 0
\][/tex]
This equation holds when \( x + 4 = 0 \). Solving \( x + 4 = 0 \), we get:
[tex]\[
x = -4
\][/tex]
This gives a repeated intercept at \( (-4,0) \).
After analyzing each function, we see that the function \( f(x) = x(x - 4) \) has the [tex]$x$[/tex]-intercepts at [tex]$(0,0)$[/tex] and [tex]$(4,0)$[/tex]. Therefore, the correct function is:
[tex]\[
\boxed{1}
\][/tex]
