To find the [tex]$x$[/tex]-intercept of an exponential function, we need to determine the value of [tex]$x$[/tex] where the function equals zero, i.e., solve [tex]$f(x) = 0$[/tex].
Let's go through each function step-by-step:
### Option A: [tex]\( f(x) = 100^{x-5} - 1 \)[/tex]
1. Set the function equal to zero:
[tex]\[ 100^{x-5} - 1 = 0 \][/tex]
2. Solve for [tex]\( 100^{x-5} \)[/tex]:
[tex]\[ 100^{x-5} = 1 \][/tex]
3. Since [tex]\( 100^0 = 1 \)[/tex]:
[tex]\[ x - 5 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
So, this function has an [tex]$x$[/tex]-intercept at [tex]\( x = 5 \)[/tex].
### Option B: [tex]\( f(x) = 3^{x-4} + 2 \)[/tex]
1. Set the function equal to zero:
[tex]\[ 3^{x-4} + 2 = 0 \][/tex]
2. Solve for [tex]\( 3^{x-4} \)[/tex]:
[tex]\[ 3^{x-4} = -2 \][/tex]
Since [tex]\( 3^{x-4} \)[/tex] is an exponential function with a positive base, it can never be negative (i.e., 3 raised to any real number will always be positive). Thus, there is no real solution to this equation.
### Option C: [tex]\( f(x) = 7^{x-1} + 1 \)[/tex]
1. Set the function equal to zero:
[tex]\[ 7^{x-1} + 1 = 0 \][/tex]
2. Solve for [tex]\( 7^{x-1} \)[/tex]:
[tex]\[ 7^{x-1} = -1 \][/tex]
Similar to option B, [tex]\( 7^{x-1} \)[/tex] is an exponential function with a positive base and cannot be negative. Therefore, there is no real solution to this equation.
### Option D: [tex]\( f(x) = -8^{x+1} - 3 \)[/tex]
1. Set the function equal to zero:
[tex]\[ -8^{x+1} - 3 = 0 \][/tex]
2. Solve for [tex]\( -8^{x+1} \)[/tex]:
[tex]\[ -8^{x+1} = 3 \][/tex]
[tex]\[ 8^{x+1} = -3 \][/tex]
Again, [tex]\( 8^{x+1} \)[/tex] with a positive base cannot be negative. Thus, there is no real solution to this equation.
### Conclusion
Only option A, [tex]\( f(x) = 100^{x-5} - 1 \)[/tex], has an [tex]$x$[/tex]-intercept at [tex]\( x = 5 \)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
