Question

erstanding graphs of exponential functions
the graph of $y = b^x$ is shown. find $b$
$b = \square$

Explanation:

Step1: Identify a point on the graph

From the graph, we can see that the curve passes through the point \((0, 1)\)? Wait, no, looking at the y - axis intersection, when \(x = 0\), the y - value is 1? Wait, no, the graph intersects the y - axis at \((0,1)\)? Wait, no, let's check the grid. Wait, the graph of \(y = b^{x}\) passes through \((0,1)\) for any \(b>0,b
eq1\). But also, let's find another point. Wait, when \(x=- 2\), what's the y - value? Wait, no, looking at the graph, when \(x = 0\), \(y = 1\)? Wait, no, the graph at \(x = 0\) seems to be at \(y = 1\)? Wait, no, maybe I made a mistake. Wait, the function is \(y=b^{x}\). Let's find a point on the graph. Let's see, when \(x = 0\), \(y = 1\) (since \(b^{0}=1\) for \(b>0,b
eq1\)). Another point: let's see, when \(x=-2\), what's \(y\)? Wait, no, looking at the graph, when \(x = 0\), \(y = 1\), and when \(x=-2\), \(y = 4\)? Wait, let's check the grid. The x - axis has marks at - 2, 0, 4. The y - axis has marks at 2, 4, 6. Wait, the graph passes through \((-2,4)\) and \((0,1)\)? Wait, let's use the point \((-2,4)\) in the equation \(y = b^{x}\). So when \(x=-2\), \(y = 4\), so \(4=b^{-2}\).

Step2: Solve for \(b\)

We know that \(b^{-2}=\frac{1}{b^{2}}\), so \(\frac{1}{b^{2}} = 4\). Then \(b^{2}=\frac{1}{4}\), so \(b=\pm\frac{1}{2}\). But since \(b>0\) (because the base of an exponential function \(y = b^{x}\) must be positive and not equal to 1), we take \(b=\frac{1}{2}\). Let's verify with \(x = 0\), \(y=(\frac{1}{2})^{0}=1\), which matches the y - intercept. When \(x=-2\), \(y = (\frac{1}{2})^{-2}=2^{2}=4\), which also matches the point \((-2,4)\) on the graph.

Answer:

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