To determine which of the given exponential functions has an [tex]\(x\)[/tex]-intercept, we need to find where these functions equal zero, that is, where [tex]\(f(x) = 0\)[/tex].
Let's analyze each function step by step:
Option A: [tex]\( f(x) = 100^{x-5} - 1 \)[/tex]
1. Set [tex]\(f(x) = 0\)[/tex]:
[tex]\[ 100^{x-5} - 1 = 0 \][/tex]
2. Isolate the exponential term:
[tex]\[ 100^{x-5} = 1 \][/tex]
3. Recognize that [tex]\(100^0 = 1\)[/tex]:
[tex]\[ x - 5 = 0 \][/tex]
4. Solve for [tex]\(x\)[/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 [tex]\(f(x) = 0\)[/tex]:
[tex]\[ 3^{x-4} + 2 = 0 \][/tex]
2. Isolate the exponential term:
[tex]\[ 3^{x-4} = -2 \][/tex]
3. Notice that [tex]\(3^{x-4}\)[/tex] is always positive for any real [tex]\(x\)[/tex], so it can never be equal to [tex]\(-2\)[/tex]:
There is no real value of [tex]\(x\)[/tex] that satisfies this equation.
Thus, this function does not have an [tex]\(x\)[/tex]-intercept.
Option C: [tex]\( f(x) = 7^{x-1} + 1 \)[/tex]
1. Set [tex]\(f(x) = 0\)[/tex]:
[tex]\[ 7^{x-1} + 1 = 0 \][/tex]
2. Isolate the exponential term:
[tex]\[ 7^{x-1} = -1 \][/tex]
3. Notice that [tex]\(7^{x-1}\)[/tex] is always positive for any real [tex]\(x\)[/tex], so it can never be equal to [tex]\(-1\)[/tex]:
There is no real value of [tex]\(x\)[/tex] that satisfies this equation.
Thus, this function does not have an [tex]\(x\)[/tex]-intercept.
Option D: [tex]\( f(x) = -8^{x+1} - 3 \)[/tex]
1. Set [tex]\(f(x) = 0\)[/tex]:
[tex]\[ -8^{x+1} - 3 = 0 \][/tex]
2. Isolate the exponential term:
[tex]\[ -8^{x+1} = 3 \][/tex]
3. Notice that [tex]\(-8^{x+1}\)[/tex] is always negative for any real [tex]\(x\)[/tex], so it can never be equal to [tex]\(3\)[/tex]:
There is no real value of [tex]\(x\)[/tex] that satisfies this equation.
Thus, this function does not have an [tex]\(x\)[/tex]-intercept.
Based on the analysis, the only function with an [tex]\(x\)[/tex]-intercept is:
A. [tex]\( f(x) = 100^{x-5} - 1 \)[/tex]
