Question

below is the graph of $y = b^x$. find $b$.
$b = \square$

Explanation:

Step1: Identify a point on the graph

From the graph, when \( x = -2 \), \( y = 6 \)? Wait, no, let's check again. Wait, maybe a better point. Wait, when \( x = 0 \), \( y = 1 \)? No, the graph is \( y = b^x \). Wait, maybe the graph passes through a point. Wait, looking at the grid, maybe when \( x = -1 \), what's y? Wait, no, let's see. Wait, the graph is a decreasing exponential, so \( 0 < b < 1 \). Let's find a point. Wait, maybe when \( x = -2 \), \( y = 4 \)? Wait, no, the grid: let's assume the graph passes through (-2, 4)? Wait, no, let's check the coordinates. Wait, the y-axis is at x=0. Let's see, when x = -2, the y-value is 4? Wait, no, the graph at x=-2 is at y=6? Wait, maybe I made a mistake. Wait, the function is \( y = b^x \). Let's find a point on the graph. Let's say when \( x = -1 \), \( y = 2 \)? Wait, no, let's look again. Wait, the graph: when x=0, y=1? No, the graph crosses the y-axis at (0,1)? Wait, no, the graph in the picture: the y-intercept is at (0,1)? Wait, no, the purple curve: when x=0, y=1? But the graph shown has a y-intercept at (0,1)? Wait, no, maybe the graph passes through (-2, 4) and (0,1)? Wait, no, let's think again. Wait, the function is \( y = b^x \). Let's take a point. Suppose when \( x = -1 \), \( y = 2 \). Then \( 2 = b^{-1} \), so \( b = \frac{1}{2} \). But that doesn't fit. Wait, maybe the point is (-2, 4). Then \( 4 = b^{-2} \), so \( b^2 = \frac{1}{4} \), so \( b = \frac{1}{2} \). Wait, no, \( b^{-2} = \frac{1}{b^2} = 4 \), so \( b^2 = \frac{1}{4} \), so \( b = \frac{1}{2} \). But let's check another point. If \( b = \frac{1}{2} \), then \( y = (\frac{1}{2})^x \). When \( x = -2 \), \( y = (\frac{1}{2})^{-2} = 4 \), which matches. When \( x = -1 \), \( y = 2 \), which also matches. So that works. Wait, but the graph in the picture: when x=-2, y=4? Let's confirm. The grid: each square is 1 unit. So at x=-2, the y-coordinate is 4. So the point (-2, 4) is on the graph.

Step2: Substitute the point into the equation

We have \( y = b^x \), and the point (-2, 4). Substitute \( x = -2 \) and \( y = 4 \) into the equation:
\( 4 = b^{-2} \)
Which is equivalent to \( 4 = \frac{1}{b^2} \)
Multiply both sides by \( b^2 \): \( 4b^2 = 1 \)
Divide both sides by 4: \( b^2 = \frac{1}{4} \)
Take the square root: \( b = \frac{1}{2} \) (since \( 0 < b < 1 \) for a decreasing exponential)

Answer:

\( \frac{1}{2} \)