To determine which function has only one [tex]\( x \)[/tex]-intercept at the point [tex]\((-6,0)\)[/tex], let’s analyze each given function by setting them equal to zero and solving for [tex]\( x \)[/tex].
### Function 1: [tex]\( f(x) = x(x - 6) \)[/tex]
Setting [tex]\( f(x) \)[/tex] to zero:
[tex]\[ x(x - 6) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \quad \text{or} \quad x = 6 \][/tex]
This function has two [tex]\( x \)[/tex]-intercepts at [tex]\( (0,0) \)[/tex] and [tex]\( (6,0) \)[/tex].
### Function 2: [tex]\( f(x) = (x - 6)(x - 6) \)[/tex]
Setting [tex]\( f(x) \)[/tex] to zero:
[tex]\[ (x - 6)(x - 6) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 6 \][/tex]
This function has one [tex]\( x \)[/tex]-intercept at [tex]\( (6,0) \)[/tex].
### Function 3: [tex]\( f(x) = (x + 6)(x - 6) \)[/tex]
Setting [tex]\( f(x) \)[/tex] to zero:
[tex]\[ (x + 6)(x - 6) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -6 \quad \text{or} \quad x = 6 \][/tex]
This function has two [tex]\( x \)[/tex]-intercepts at [tex]\( (-6,0) \)[/tex] and [tex]\( (6,0) \)[/tex].
### Function 4: [tex]\( f(x) = (x + 6)(x + 6) \)[/tex]
Setting [tex]\( f(x) \)[/tex] to zero:
[tex]\[ (x + 6)(x + 6) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -6 \][/tex]
This function has one [tex]\( x \)[/tex]-intercept at [tex]\( (-6,0) \)[/tex].
From the analysis:
- Function 1 has intercepts at [tex]\( (0,0) \)[/tex] and [tex]\( (6,0) \)[/tex]
- Function 2 has an intercept at [tex]\( (6,0) \)[/tex]
- Function 3 has intercepts at [tex]\( (-6,0) \)[/tex] and [tex]\( (6,0) \)[/tex]
- Function 4 has an intercept at [tex]\( (-6,0) \)[/tex]
Therefore, the function that has only one [tex]\( x \)[/tex]-intercept at [tex]\( (-6,0) \)[/tex] is:
[tex]\[ \boxed{f(x) = (x + 6)(x + 6)} \][/tex]
Hence, the answer is:
[tex]\[ \boxed{4} \][/tex]
