Exponential Functions: Finding The X-intercept

Hey everyone! Today, we're diving into the fascinating world of exponential functions and tackling a super common question: which exponential function actually has an x-intercept? This might sound a little tricky at first, but trust me, guys, once you get the hang of it, it's a piece of cake. We'll break down the concept, look at the options, and figure out which one hits the mark. So, grab your calculators, get comfy, and let's get this math party started! Understanding the core properties of exponential functions is key here. Remember, an exponential function generally takes the form $f(x) = a imes b^x + c$, where '$a$' is the initial value, '$b$' is the base (and must be positive and not equal to 1), and '$x$' is the exponent. The '$c$' term is a vertical shift. The x-intercept is the point where the graph crosses the x-axis. At this point, the y-value, or $f(x)$, is equal to zero. So, we're essentially looking for a function where $a imes b^x + c = 0$ has a real solution for '$x$'. The behavior of exponential functions is heavily influenced by the base '$b$' and the vertical shift '$c$'. If the base '$b$' is greater than 1, the function will increase as '$x$' increases. If '$b$' is between 0 and 1, the function will decrease as '$x$' increases. The vertical shift '$c$' moves the entire graph up or down. This shift is crucial because it determines whether the function's range overlaps with the x-axis (where $y=0$). The horizontal asymptote of an exponential function $f(x) = a imes b^x + c$ is $y=c$. This means the function's values get arbitrarily close to '$c$' as '$x$' approaches positive or negative infinity, but they never actually reach '$c$' (unless $a=0$, which isn't a typical exponential function). For an x-intercept to exist, the function must be able to reach a y-value of 0. This means that the range of the function must include 0. The range of $f(x) = a imes b^x + c$ depends on the sign of '$a$' and the horizontal asymptote '$c$'. If '$a$' is positive, the range is $(c, o)$ if $b > 1$ or $( o, c)$ if $0 < b < 1$. If '$a$' is negative, the range is $( o, c)$ if $b > 1$ or $(c, o)$ if $0 < b < 1$. In simpler terms, if the function is always positive and its horizontal asymptote is negative, it might cross the x-axis. Conversely, if the function is always negative and its horizontal asymptote is positive, it might also cross the x-axis. Let's keep these ideas in mind as we examine each option. The visual representation of these functions is also super helpful. Think about graphs: exponential functions usually have a curve that either goes up steeply or down steeply, approaching a horizontal line. The x-intercept is just where that curve kisses the horizontal x-axis. It's all about where the graph is positioned vertically. So, the critical factor is whether the function can actually produce a zero output, given its inherent behavior and any vertical shifts applied. This involves understanding the limits and the range of the function. We're looking for a function whose range includes zero.

Analyzing the Options: Which Function Has an x-intercept?

Alright, guys, let's roll up our sleeves and dissect each of the given exponential functions. The core idea is to determine if any of these functions can actually output a value of zero for $f(x)$. Remember, an x-intercept occurs when $f(x) = 0$. So, for each option, we're essentially asking: can $a imes b^x + c = 0$ be solved for '$x$'?

Option A: $f(x)=100^{x-5}-1$

First up, we have $f(x)=100^{x-5}-1$. Here, our base is 100, which is greater than 1, so this function will increase as '$x$' increases. The '$a$' term is implicitly 1 (since there's no number multiplying the exponential term), and it's positive. The vertical shift is '-1', meaning the horizontal asymptote is at $y=-1$. Since the base is greater than 1 and '$a$' is positive, the function's values will always be greater than the horizontal asymptote. In this case, $f(x) > -1$. Can this function ever equal zero? Let's set it to zero: $100^{x-5}-1 = 0$. If we add 1 to both sides, we get $100^{x-5} = 1$. Now, we know that any non-zero number raised to the power of 0 equals 1. So, we can set the exponent equal to 0: $x-5 = 0$. Solving for '$x$' gives us $x=5$. So, yes! This function does have an x-intercept at $(5, 0)$. This is a strong contender, folks! The function's graph starts below the x-axis (approaching $y=-1$) and then increases, eventually crossing the x-axis at $x=5$. The key here is that the horizontal asymptote ($y=-1$) is below the x-axis ($y=0$), and the function increases, meaning it must cross the x-axis to get from values close to -1 to positive values. The range of this function is $(-1, o)$, which clearly includes 0.

Option B: $f(x)=3^{x-4}+2$

Next, let's look at $f(x)=3^{x-4}+2$. The base here is 3 (greater than 1), '$a$' is positive (implicitly 1), and the vertical shift is '+2'. This means the horizontal asymptote is at $y=2$. Since '$a$' is positive and the base is greater than 1, the function's values will always be greater than the horizontal asymptote. So, $f(x) > 2$. Can this function ever equal zero? If we set $f(x)=0$, we get $3^{x-4}+2 = 0$. Subtracting 2 from both sides gives us $3^{x-4} = -2$. Here's the crucial part, guys: a positive base (like 3) raised to any real power will always result in a positive number. It can never be negative. Therefore, there is no real value of '$x$' that can satisfy $3^{x-4} = -2$. This means this function does not have an x-intercept. Its entire graph lies above the horizontal asymptote $y=2$, and thus, it never touches or crosses the x-axis. The range is $(2, o)$.

Option C: $f(x)=7^{x-1}+1$

Moving on to option C: $f(x)=7^{x-1}+1$. The base is 7 (greater than 1), '$a$' is positive (implicitly 1), and the vertical shift is '+1'. The horizontal asymptote is therefore at $y=1$. Similar to option B, since '$a$' is positive and the base is greater than 1, the function's values will always be greater than the horizontal asymptote. So, $f(x) > 1$. Setting $f(x)=0$ gives us $7^{x-1}+1 = 0$. Subtracting 1 from both sides yields $7^{x-1} = -1$. Again, a positive base (7) raised to any real power can only produce positive results. It can never equal -1. Thus, this function also does not have an x-intercept. The entire graph is above the line $y=1$, so it never reaches $y=0$. The range is $(1, o)$.

Option D: $f(x)=-8^{x+1}-3$

Finally, let's analyze option D: $f(x)=-8^{x+1}-3$. The base is 8 (greater than 1). However, notice the negative sign in front of the exponential term: '$a$' is implicitly -1, which is negative. The vertical shift is '-3'. This means the horizontal asymptote is at $y=-3$. Because '$a$' is negative, the function's values will always be less than the horizontal asymptote. So, $f(x) < -3$. Can this function equal zero? Setting $f(x)=0$ gives us $-8^{x+1}-3 = 0$. Adding 3 to both sides gives $-8^{x+1} = 3$. Multiplying by -1, we get $8^{x+1} = -3$. Once again, we have a positive base (8) raised to a power equaling a negative number. This is impossible for any real value of '$x$'. Therefore, this function does not have an x-intercept. The entire graph lies below the line $y=-3$, so it never reaches $y=0$. The range is $( o, -3)$.

The Verdict: Which Function Has an x-intercept?

After meticulously examining each option, it's clear that only one function satisfies the condition for having an x-intercept. Let's recap our findings:

• Option A: $f(x)=100^{x-5}-1$: Has a horizontal asymptote at $y=-1$ and the function increases. Since the asymptote is below the x-axis and the function increases, it must cross the x-axis. We found the intercept at $x=5$.
• Option B: $f(x)=3^{x-4}+2$: Has a horizontal asymptote at $y=2$. Since the function is always above this asymptote (range is $(2, o)$), it never reaches $y=0$.
• Option C: $f(x)=7^{x-1}+1$: Has a horizontal asymptote at $y=1$. The function is always above this asymptote (range is $(1, o)$), so it never reaches $y=0$.
• Option D: $f(x)=-8^{x+1}-3$: Has a horizontal asymptote at $y=-3$. Since the leading coefficient is negative, the function is always below this asymptote (range is $( o, -3)$), so it never reaches $y=0$.

Therefore, the correct answer is A. This is because the function's behavior (increasing) and its vertical shift combine to allow it to reach and cross the x-axis. It's all about that horizontal asymptote being on the