Question

y = e^{x}+5 y = e^{x}+4 y = \ln x+4 y = \ln x+5

Explanation:

Step1: Analyze the domain of the function

The domain of \(y = \ln x\) is \(x>0\). In the given graph, when \(x = 0\), the function is defined. So we can rule out the functions with \(\ln x\) (because \(\ln x\) is not defined at \(x = 0\)). So \(y=\ln x + 4\) and \(y=\ln x+5\) are not correct.

Step2: Substitute \(x = 0\) into the remaining exponential - function options

For \(y=e^{x}+5\), when \(x = 0\), \(y=e^{0}+5=1 + 5=6\). For \(y=e^{x}+4\), when \(x = 0\), \(y=e^{0}+4=1 + 4=5\). Looking at the graph, when \(x = 0\), \(y\) is around \(2\). Wait, there is a mistake. Let's recall the general form of \(y = e^{x}\) has \(y(0)=1\). If we assume the \(y\) - intercept of the given graph is \(y(0)\approx2\). Wait, no, looking at the options again. Wait, another approach: the general shape of \(y = e^{x}\) (exponential growth) is correct for the graph (since \(\ln x\) has a different shape, for \(y = \ln x\), as \(x\to0^{+}\), \(y\to-\infty\)). And if we check the \(y\) - intercept. Let's assume \(y = e^{x}+4\). When \(x = 0\), \(y=e^{0}+4=1 + 4 = 5\) (not correct). Wait, maybe the original \(y = e^{x}\) has \(y(0)=1\). If we assume the graph has \(y\) - intercept \(y(0)\approx2\). Wait, no, looking at the options again. Wait, another thought: the function \(y = e^{x}+4\) is an exponential function \(y = e^{x}\) shifted up by \(4\) units. The function \(y=\ln x+4\) is a logarithmic function. Since the graph is defined for \(x = 0\) (logarithmic functions are not defined at \(x = 0\)), and the shape is of an exponential function (starts to increase rapidly and then levels off).

Answer:

\(y = e^{x}+4\)