Question

exercise 4.5.3
write an equation for the function graphed here.
$f(x) = \square$

Explanation:

Step1: Identify the function type

The graph appears to be an exponential function. The general form of an exponential function is \( f(x) = a \cdot b^x + k \), but since it has a horizontal asymptote and passes through certain points, we can assume it's a shifted exponential. Let's check the y-intercept and the point where it crosses the x-axis. At \( x = 0 \), \( y \approx -1.5 \)? Wait, no, looking at the graph, when \( x = 0 \), \( y = -1.5 \)? Wait, actually, let's see the key points. The graph crosses the x-axis at \( x = 2 \), so \( f(2) = 0 \). Also, as \( x \to -\infty \), \( f(x) \to -2 \), so the horizontal asymptote is \( y = -2 \). So the general form for a shifted exponential is \( f(x) = a \cdot b^x - 2 \).

Step2: Find the value of \( a \) and \( b \)

We know that when \( x = 2 \), \( f(2) = 0 \). Also, let's check the y-intercept. When \( x = 0 \), \( f(0) = -1.5 \)? Wait, no, looking at the graph, at \( x = 0 \), the y-value is -1.5? Wait, maybe the horizontal asymptote is \( y = -2 \), so the function is \( f(x) = a \cdot b^x - 2 \). Let's use the point \( (2, 0) \) and \( (0, -1.5) \)? Wait, no, maybe it's a simpler exponential. Wait, another approach: the standard exponential function \( f(x) = b^x \) shifted. Wait, maybe the function is \( f(x) = 2^{x - 2} - 2 \)? Wait, let's test \( x = 2 \): \( 2^{0} - 2 = 1 - 2 = -1 \), no. Wait, maybe \( f(x) = 2^{x - 2} - 2 \) is not right. Wait, let's check the point \( (2, 0) \). Let's assume the function is \( f(x) = a \cdot 2^x + k \). Wait, the horizontal asymptote is \( y = -2 \), so \( k = -2 \). So \( f(x) = a \cdot 2^x - 2 \). Now, when \( x = 2 \), \( f(2) = 0 \): \( a \cdot 2^2 - 2 = 0 \implies 4a - 2 = 0 \implies 4a = 2 \implies a = 0.5 \). So \( f(x) = 0.5 \cdot 2^x - 2 \). Simplify \( 0.5 \cdot 2^x = 2^{x - 1} \), so \( f(x) = 2^{x - 1} - 2 \). Let's check \( x = 0 \): \( 2^{-1} - 2 = 0.5 - 2 = -1.5 \), which matches the y-intercept. \( x = 2 \): \( 2^{1} - 2 = 2 - 2 = 0 \), which matches the x-intercept. \( x = 3 \): \( 2^{2} - 2 = 4 - 2 = 2 \), which looks like the graph at \( x = 3 \) is around 2, which matches. So the function is \( f(x) = 2^{x - 1} - 2 \) or equivalently \( f(x) = \frac{1}{2} \cdot 2^x - 2 = 2^{x - 1} - 2 \).

Answer:

\( f(x) = 2^{x - 1} - 2 \) (or equivalent forms like \( f(x) = \frac{1}{2} \cdot 2^x - 2 \) or \( f(x) = 2^{x - 1} - 2 \))