Which description matches the graph of $f\left(x\right)=\log x+3$ (base 10)?

• A. An exponential curve shifted up 3 units, with horizontal asymptote $y=3$, and passing through $(0,4)$.
• B. A logarithmic curve shifted right 3 units, with vertical asymptote $x=3$, and passing through $(4,0)$.
• C. A logarithmic curve shifted up 3 units, with vertical asymptote $x=0$, domain $x>0$, and passing through $(1,3)$.
• D. A logarithmic curve shifted down 3 units, with vertical asymptote $x=0$, and passing through $(1,-3)$.

Answer: (C)

Verified step by step guidance

1. Identify the base function: The given function is \(f(x) = \log x + 3\), which is a logarithmic function with base 10, shifted vertically.
2. Understand the vertical shift: The \(+3\) outside the logarithm means the entire graph of \(\log x\) is shifted up by 3 units.
3. Recall the vertical asymptote of \(\log x\): The function \(\log x\) has a vertical asymptote at \(x=0\), since \(\log x\) is undefined for \(x \leq 0\).
4. Determine the domain: Since \(\log x\) is only defined for \(x > 0\), the domain of \(f(x)\) remains \(x > 0\) even after the vertical shift.
5. Find a point on the graph: Evaluate \(f(1) = \log 1 + 3 = 0 + 3 = 3\), so the graph passes through the point \((1, 3)\).