The question asks which graph represents the function $y = \log_2 x$.

Key characteristics of the logarithmic function $y = \log_2 x$:

1. Domain: The function is only defined for $x > 0$, so the graph must be entirely to the right of the y-axis.
2. Vertical asymptote: There is a vertical asymptote at $x = 0$, meaning the curve approaches but never touches or crosses the y-axis.
3. Shape: The logarithmic graph increases slowly for larger values of $x$ but rapidly decreases near 0 on the positive x-axis.
4. Intercept: The graph will pass through $(1, 0)$ because $\log_2(1) = 0$.

Looking at the four graphs:

• Top left: This graph is defined for negative x-values and approaches the x-axis from the left, which doesn't match a logarithmic function.
• Top right: This is an exponential curve, not a logarithmic curve.
• Bottom left: This is also incorrect as it represents a negative logarithmic-like shape.
• Bottom right: This graph fits the description of $y = \log_2 x$, defined only for $x > 0$, with a vertical asymptote at $x = 0$, and passing through $(1, 0)$.

Thus, the correct graph is the bottom right one.

Would you like more details or have any questions?

Here are five related questions to explore:

1. How does the base of a logarithm affect the shape of its graph?
2. What is the inverse of the function $y = \log_2 x$?
3. Can you explain the domain and range of logarithmic functions?
4. How would the graph of $y = \log_3 x$ compare to $y = \log_2 x$?
5. What happens to the graph of $y = \log_2 x$ if you add a constant to the function (e.g., $y = \log_2 x + 2$)?

Tip: When sketching logarithmic graphs, always check for the vertical asymptote at $x = 0$ and where the function crosses the x-axis (at $(1, 0)$).