Question

which table represents the graph of a logarithmic function with both an (x)-and (y)-intercept?

Explanation:

Analyze the requirements of the function

Using the Logarithmic Functions and Logarithmic Intercepts knowledge points

We are looking for a logarithmic function of the form \(y = a \log_b(x - h) + k\) that has:

1. An \(x\)-intercept: The graph crosses the \(x\)-axis, meaning there is a point where \(y = 0\).
2. A \(y\)-intercept: The graph crosses the \(y\)-axis, meaning \(x = 0\) must be in the domain of the function. For a logarithmic function \(y = a \log_b(x - h) + k\), this requires the vertical asymptote \(x = h\) to be to the left of the \(y\)-axis, so \(h < 0\). Thus, the domain \(x > h\) includes \(x = 0\).

Evaluate the first table

Using the Logarithmic Functions knowledge point

Let's examine the first table:

• Points: \((3, 0)\), \((4, -15)\), \((5, 0.585)\), \((6, 1.322)\), \((7, 1.807)\)
• The value \(y = -15\) at \(x = 4\) is highly anomalous and does not fit a smooth logarithmic curve passing through \((3, 0)\), \((5, 0.585)\), \((6, 1.322)\), and \((7, 1.807)\).
• Furthermore, if \(x = 3\) gives \(y = 0\), and the values for \(x \ge 5\) grow slowly, a standard logarithm would not drop sharply to \(-15\) at \(x = 4\) and then rise back up. This table does not represent a standard logarithmic function.

Evaluate the second table

Using the Logarithmic Functions and Logarithmic Intercepts knowledge points

Let's examine the second table:

• Points: \((-1.5, -1)\), \((-0.5, 0.585)\), \((0.5, 1.322)\), \((1.5, 1.807)\), \((2.5, 2.169)\)
• Notice the \(x\)-values are shifted by \(2\) compared to a standard table. Let's test a transformed logarithmic function of the form:

$$y = \log_2(x + 2)$$

• For \(x = -1.5\): \(y = \log_2(-1.5 + 2) = \log_2(0.5) = -1\) (Matches!)
• For \(x = -0.5\): \(y = \log_2(-0.5 + 2) = \log_2(1.5) \approx 0.585\) (Matches!)
• For \(x = 0.5\): \(y = \log_2(0.5 + 2) = \log_2(2.5) \approx 1.322\) (Matches!)
• For \(x = 1.5\): \(y = \log_2(1.5 + 2) = \log_2(3.5) \approx 1.807\) (Matches!)
• For \(x = 2.5\): \(y = \log_2(2.5 + 2) = \log_2(4.5) \approx 2.169\) (Matches!)
• Since the domain is \(x > -2\), \(x = 0\) is in the domain, which means it has a \(y\)-intercept:

$$y = \log_2(0 + 2) = 1$$

• Since the range of any logarithmic function is \((-\infty, \infty)\), it must cross \(y = 0\), which gives the \(x\)-intercept:

$$0 = \log_2(x + 2) \implies x + 2 = 1 \implies x = -1$$

• Thus, this table perfectly represents a logarithmic function with both intercepts.

Evaluate the third table

Using the Logarithmic Functions knowledge point

Let's examine the third table:

• Points: \((0.5, -0.631)\), \((1.5, 0.369)\), \((2.5, 0.834)\), \((3.5, 1.146)\), \((4.5, 1.369)\)
• Let's test a logarithmic function of the form \(y = \log_b(x - h) + k\).
• The differences in \(y\) values do not align with a domain that extends to include \(x = 0\) while maintaining both intercepts in the same clean manner as the second table.

Answer:

• (A)

$$\begin{array}{|c|c|} \hline x & y \\ \hline 3 & 0 \\ \hline 4 & -15 \\ \hline 5 & 0.585 \\ \hline 6 & 1.322 \\ \hline 7 & 1.807 \\ \hline \end{array}$$

• **(B)

$$\begin{array}{|c|c|} \hline x & y \\ \hline -1.5 & -1 \\ \hline -0.5 & 0.585 \\ \hline 0.5 & 1.322 \\ \hline 1.5 & 1.807 \\ \hline 2.5 & 2.169 \\ \hline \end{array}$$

(Correct answer)**

• (C)

$$\begin{array}{|c|c|} \hline x & y \\ \hline 0.5 & -0.631 \\ \hline 1.5 & 0.369 \\ \hline 2.5 & 0.834 \\ \hline 3.5 & 1.146 \\ \hline 4.5 & 1.369 \\ \hline \end{array}$$