Question

¿qué función está graficada?
$y = 2^{x - 3} + 2$
$y = 2^{x + 3} - 2$
$y = 2^{x - 2} + 3$

Explanation:

Step1: Analizar la asintota horizontal

La gráfica tiene una asintota horizontal en \( y = 3 \)? No, en la gráfica se ve que la curva se acerca a \( y = 3 \)? Espera, revisemos la gráfica. La curva en el eje \( y \) (cuando \( x = 0 \)): vamos a probar \( x = 0 \) en cada función.

Step2: Probar \( x = 0 \) en cada función

• Para \( y = 2^{x - 3} + 2 \): \( y = 2^{-3} + 2 = \frac{1}{8} + 2 \approx 2.125 \)
• Para \( y = 2^{x + 3} - 2 \): \( y = 2^{3} - 2 = 8 - 2 = 6 \) (no coincide, la gráfica en \( x = 0 \) está cerca de 3? Espera, la tercera función: \( y = 2^{x - 2} + 3 \). Cuando \( x = 2 \), \( y = 2^{0} + 3 = 1 + 3 = 4 \). Wait, la gráfica en \( x = 4 \) parece alcanzar 6? Wait, la tercera opción: \( y = 2^{x - 2} + 3 \). Cuando \( x = 2 \), \( y = 1 + 3 = 4 \). Cuando \( x = 4 \), \( y = 2^{2} + 3 = 4 + 3 = 7 \)? No, la gráfica en \( x = 4 \) está en 6? Wait, quizás me equivoqué. Wait, la gráfica: la curva en \( x = 2 \) es \( y = 4 \)? Wait, la tercera función: \( y = 2^{x - 2} + 3 \). Let's check the horizontal asymptote. The parent function \( 2^x \) has horizontal asymptote \( y = 0 \). For \( y = 2^{x - 2} + 3 \), the horizontal asymptote is \( y = 3 \). Looking at the graph, the curve approaches \( y = 3 \) as \( x \to -\infty \), which matches. Now check a point: when \( x = 2 \), \( y = 2^{0} + 3 = 4 \), which is on the graph (since at \( x = 2 \), the graph is at \( y = 4 \)? Wait, the graph: when \( x = 4 \), \( y = 2^{4 - 2} + 3 = 4 + 3 = 7 \)? No, the graph at \( x = 4 \) is at 6? Wait, maybe I made a mistake. Wait, the third option is \( y = 2^{x - 2} + 3 \). Let's check the first option: \( y = 2^{x - 3} + 2 \). Horizontal asymptote \( y = 2 \). The graph's horizontal asymptote: looking at the graph, as \( x \to -\infty \), the curve approaches \( y = 3 \)? No, the graph shows that as \( x \) decreases, the curve approaches a horizontal line. Let's look at the y-intercept (x=0). The graph at x=0: let's see the grid. The y-axis: the curve is at y=3? Wait, the third function: \( y = 2^{0 - 2} + 3 = 2^{-2} + 3 = 0.25 + 3 = 3.25 \), which is close. Wait, maybe the correct function is \( y = 2^{x - 2} + 3 \). Let's verify:

• Horizontal asymptote: \( y = 3 \) (since \( 2^{x - 2} \to 0 \) as \( x \to -\infty \), so \( y \to 3 \)), which matches the graph (the left end approaches \( y = 3 \)).
• When \( x = 2 \), \( y = 2^{0} + 3 = 4 \), which is on the graph (the curve passes through (2, 4)).
• When \( x = 4 \), \( y = 2^{2} + 3 = 7 \)? No, the graph at \( x = 4 \) is at 6? Wait, maybe the graph's x=4 is at 6, but maybe my estimation is wrong. Alternatively, let's check the other options:

First option: \( y = 2^{x - 3} + 2 \). Horizontal asymptote \( y = 2 \). The graph's left end is above 2, but closer to 3? No, the third option's asymptote is 3, which matches the graph's left end (approaching 3). Second option: \( y = 2^{x + 3} - 2 \). Horizontal asymptote \( y = -2 \), which doesn't match. So the correct function is \( y = 2^{x - 2} + 3 \).

Answer:

\( y = 2^{x - 2} + 3 \) (la tercera opción: \( y = 2^{x - 2} + 3 \))