Question

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

Explanation:

Step1: Analyze the function \(y = \ln x\)

The domain of \(y=\ln x\) is \(x>0\). But the given graph has points for \(x = 0\) (not in the domain of \(y=\ln x\) and \(y=\ln x + 1\) as their domains are also \(x>0\)). So, \(y=\ln x\) and \(y=\ln x + 1\) can be excluded.

Step2: Analyze the function \(y = e^{x}+1\)

When \(x = 0\), \(y=e^{0}+1=1 + 1=2
eq1\).

Step3: Analyze the function \(y = e^{x}\)

When \(x = 0\), \(y=e^{0}=1\). Also, the general shape of \(y = e^{x}\) (exponential growth) matches the given graph.

Answer:

\(y = e^{x}\)