Question

which equation is represented by the graph below?

\\(y = \ln x\\)
\\(y = \ln x + 1\\)
\\(y = e^x\\)

Explanation:

⚡ Using what you learned: Graphing Exponential Functions

Step 1: Identify key features of the graph

Observe the behavior of the graph:

• As \( x \to -\infty \), the graph approaches the horizontal line \( y = 0 \) (the x-axis). This indicates a horizontal asymptote at \( y = 0 \).
• The graph passes through the y-intercept at \( (0, 1) \).
• As \( x \) increases, the graph grows rapidly upward.
• The domain of the function is all real numbers, \( (-\infty, \infty) \).

Step 2: Evaluate the given options

Let's test the given equations against these features:

1. Option 1: \( y = \ln x \)

• The natural logarithmic function is only defined for \( x > 0 \). Its graph has a vertical asymptote at \( x = 0 \) (the y-axis) and passes through \( (1, 0) \). This does not match the given graph.

1. Option 2: \( y = \ln x + 1 \)

• Similar to \( y = \ln x \), this logarithmic function is only defined for \( x > 0 \) and has a vertical asymptote at \( x = 0 \). This does not match the given graph.

1. Option 3: \( y = e^x \)

• This is a natural exponential function.
• Its domain is all real numbers.
• It has a horizontal asymptote at \( y = 0 \) as \( x \to -\infty \).
• When \( x = 0 \), \( y = e^0 = 1 \), giving a y-intercept of \( (0, 1) \).
• When \( x = 1 \), \( y = e^1 \approx 2.718 \), which matches the point plotted on the graph.

Answer:

\( y = e^x \)