Question

1. which is the graph of the function $y = -3(2)^x$? (with two graph options shown)

Explanation:

Step 1: Analyze the function type

The function \( y = a(b)^x \) is an exponential function. If \( a>0 \) and \( b > 1 \), it's an exponential growth function.

Step 2: Find the y - intercept

For \( y=a(2)^x \), when \( x = 0 \), \( y=a \). If we assume \( a = 2 \) (matching the first graph's y - intercept at \( y = 2 \) when \( x = 0 \)), then the function is \( y = 2(2)^x \).

Step 3: Analyze the shape of the graph

As \( x\) increases, \( 2^x \) increases, so \( y = 2(2)^x \) increases exponentially. The first graph (top one) shows a curve that starts near the x - axis for negative \( x\), passes through \( (0,2) \), and increases rapidly for positive \( x\), which is consistent with an exponential growth function \( y = 2(2)^x \). The second graph seems to have a steeper curve, maybe for \( a = 3 \).

(Note: The original function has a negative sign which makes it a decreasing function, but the given graphs are increasing, so there is likely a typo. The above analysis is based on assuming a positive coefficient for the exponential function.)

Answer:

(Note: Since the original problem's graph options might be incomplete in the provided text, but based on the function \( y = -3(2)^x \), let's analyze:

1. Y - intercept: When \( x = 0 \), \( y=-3(2)^0=-3(1) = - 3 \). Wait, but the given graphs have y - intercepts at 2 or 3? Wait, maybe there is a typo or mis - writing. If the function is \( y = 3(2)^x \) (positive coefficient), then at \( x = 0 \), \( y = 3(1)=3 \), and as \( x\) increases, \( y\) increases exponentially. If the function is \( y=-3(2)^x \), then at \( x = 0 \), \( y=-3 \), and as \( x\) increases, \( y\) decreases (since the base \( 2>1 \) and the coefficient is negative, it's an exponential decay towards negative infinity). But the given graphs show increasing curves, so maybe the function is \( y = 3(2)^x \) or \( y = 2(2)^x=2^{x + 1}\).

Assuming the function is \( y = 3(2)^x \) (maybe a typo in the sign), let's check the y - intercept. At \( x = 0 \), \( y = 3\). The first graph (top one) has a y - intercept at 2? Wait, no, the first graph has a y - intercept at 2? Wait, the first graph: when \( x = 0 \), the curve passes through \( (0,2) \)? No, the first graph's curve at \( x = 0 \) is at \( y = 2 \), the second at \( y = 3 \) (approx). Wait, maybe the function is \( y = 2(2)^x=2^{x + 1}\), then at \( x = 0 \), \( y = 2 \), which matches the first graph.

If we consider the function \( y=-3(2)^x \), there is no graph among the given ones (since they are increasing), so maybe the sign is wrong. If we assume the function is \( y = 3(2)^x \), then the second graph (the one with a steeper curve, maybe) or if it's \( y = 2(2)^x \), the first graph.

But since the user's problem might have a typo, but based on the given graphs (assuming the function is \( y = 2(2)^x \) or \( y = 3(2)^x \)):

If the function is \( y = 2(2)^x \), then at \( x = 0 \), \( y = 2 \), and as \( x\) increases, \( y\) increases exponentially. The first graph (the top one with the grid) shows a curve that passes through \( (0,2) \) and increases, so the answer would be "The top graph (the first one with the x - axis from - 4 to 4 and y - axis from - 2 to 8)".)

(But due to the ambiguity in the function's sign, we'll proceed with the assumption that there is a typo and the function is \( y = 2(2)^x \) or \( y = 3(2)^x \) with a positive coefficient. So the answer is the top graph (the first graph shown).)