Question

which equation is represented by the graph below?

o y = e ^ { x }
o y = e ^ { x } - 1
o y = ln x
o y = ln x - 1

Explanation:

Step1: Analyze the function \(y = e^{x}\)

The function \(y = e^{x}\) has a \(y -\)intercept at \((0,1)\) (when \(x = 0\), \(y=e^{0}=1\)). But the given graph has a \(y -\)intercept at \((0, - 2)\) approximately. So \(y = e^{x}\) is not the correct function.

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

When \(x = 0\), \(y=e^{0}-1=1 - 1=0\). The \(y -\)intercept of \(y = e^{x}-1\) is \((0,0)\). So \(y = e^{x}-1\) is not the correct function.

Step3: Analyze the function \(y=\ln x\)

The domain of \(y = \ln x\) is \(x>0\). When \(x = 1\), \(y=\ln(1) = 0\). But we can also check the transformation.

Step4: Analyze the function \(y=\ln x-1\)

We know that the graph of \(y = f(x)+k\) is a vertical shift of the graph of \(y = f(x)\). The graph of \(y=\ln x\) has a point \((1,0)\). For the function \(y=\ln x-1\), when \(x = 1\), \(y=\ln(1)-1=-1\). Also, using the property of the logarithmic function \(y = \ln x\) (domain \(x>0\), passes through \((1,0)\)) and the vertical - shift rule \(y=f(x)-1\) (shifts the graph of \(y = f(x)\) down by 1 unit).

Answer:

\(y=\ln x - 1\)