graph the logarithmic function.\n$y = 3\\log_2(2x - 1)+1$

Explanation:

Step1: Find the vertical - asymptote

Set the argument of the logarithm greater than 0. $2x - 1>0$, solving for $x$ gives $x>\frac{1}{2}$. So the vertical asymptote is $x = \frac{1}{2}$.

Step2: Find the $y$ - intercept

Set $x = 0$. But $x=0$ is not in the domain ($x>\frac{1}{2}$), so there is no $y$ - intercept.

Step3: Find the $x$ - intercept

Set $y = 0$. Then $0=3\log_2(2x - 1)+1$. First, isolate the logarithm: $- 1=3\log_2(2x - 1)$, so $\log_2(2x - 1)=-\frac{1}{3}$. Using the definition of logarithms $2^{-\frac{1}{3}}=2x - 1$, and $x=\frac{2^{-\frac{1}{3}}+1}{2}\approx\frac{\frac{1}{\sqrt[3]{2}}+1}{2}\approx\frac{0.794 + 1}{2}=0.897>\frac{1}{2}$.

Step4: Analyze the behavior

As $x\to\frac{1}{2}^+$, $y\to-\infty$. As $x\to+\infty$, since the coefficient of the logarithm is positive ($3>0$), $y\to+\infty$.

The graph of $y = 3\log_2(2x - 1)+1$ has a vertical asymptote at $x=\frac{1}{2}$, passes through the $x$ - intercept $(\frac{2^{-\frac{1}{3}}+1}{2},0)$ and increases as $x$ increases for $x>\frac{1}{2}$. Looking at the given graphs, the correct one is the one that has a vertical asymptote at $x = \frac{1}{2}$ and is an increasing curve for $x>\frac{1}{2}$.

Answer:

The graph that has a vertical asymptote at $x=\frac{1}{2}$ and is increasing for $x>\frac{1}{2}$ (the first graph in the top - left if we assume a standard left - to - right, top - to - bottom numbering of the four graphs).