To find which exponential function has an [tex]$x$[/tex]-intercept, we set each function equal to zero and solve for [tex]$x$[/tex]. An [tex]$x$[/tex]-intercept occurs where [tex]$f(x) = 0$[/tex].
Let's examine each function one by one:
### Option A: [tex]\( f(x) = 100^{x-5} - 1 \)[/tex]
To find the [tex]$x$[/tex]-intercept:
[tex]\[ 100^{x-5} - 1 = 0 \][/tex]
[tex]\[ 100^{x-5} = 1 \][/tex]
Recognize that [tex]\( 100^0 = 1 \)[/tex], so:
[tex]\[ x - 5 = 0 \][/tex]
[tex]\[ x = 5 \][/tex]
This function has an [tex]$x$[/tex]-intercept at [tex]\( x = 5 \)[/tex].
### Option B: [tex]\( f(x) = 3^{x-4} + 2 \)[/tex]
To find the [tex]$x$[/tex]-intercept:
[tex]\[ 3^{x-4} + 2 = 0 \][/tex]
[tex]\[ 3^{x-4} = -2 \][/tex]
Notice that [tex]\( 3^{x-4} \)[/tex] is always positive for all real numbers [tex]\( x \)[/tex]. It can never equal [tex]$-2$[/tex]. So, this function has no [tex]$x$[/tex]-intercept.
### Option C: [tex]\( f(x) = 7^{x-1} + 1 \)[/tex]
To find the [tex]$x$[/tex]-intercept:
[tex]\[ 7^{x-1} + 1 = 0 \][/tex]
[tex]\[ 7^{x-1} = -1 \][/tex]
Notice that [tex]\( 7^{x-1} \)[/tex] is always positive for all real numbers [tex]\( x \)[/tex]. It can never equal [tex]$-1$[/tex]. So, this function has no [tex]$x$[/tex]-intercept.
### Option D: [tex]\( f(x) = -8^{x+1} - 3 \)[/tex]
To find the [tex]$x$[/tex]-intercept:
[tex]\[ -8^{x+1} - 3 = 0 \][/tex]
[tex]\[ -8^{x+1} = 3 \][/tex]
[tex]\[ 8^{x+1} = -3 \][/tex]
Notice that [tex]\( 8^{x+1} \)[/tex] is always positive for all real numbers [tex]\( x \)[/tex]. It can never equal [tex]$-3$[/tex]. So, this function has no [tex]$x$[/tex]-intercept.
### Conclusion
Only [tex]\( f(x) = 100^{x-5} - 1 \)[/tex] (Option A) has an [tex]$x$[/tex]-intercept.
Thus, the exponential function that has an [tex]$x$[/tex]-intercept is:
[tex]\[ \boxed{A} \][/tex]
