Question

graph ( y=log _{2}(-3 x-2) ).
to graph, first select the function, then select a point on its asymptote, and finally selecting two points o
the graph.

Explanation:

Step1: Find the vertical asymptote

For a logarithmic function \(y = \log_{a}(u)\), the vertical asymptote occurs when \(u = 0\).
Set \(-3x - 2=0\).
Solve for \(x\):
\(-3x=2\), so \(x =-\frac{2}{3}\).
A point on the asymptote could be \((-\frac{2}{3},0)\) (any \(y\)-value is fine since it's a vertical line \(x =-\frac{2}{3}\)).

Step2: Find two points on the graph

Let \(y = 0\):
\(0=\log_{2}(-3x - 2)\)
By the definition of logarithms, if \(\log_{a}(b)=c\), then \(b=a^{c}\). So \(-3x - 2=2^{0}=1\)
\(-3x=1 + 2=3\), \(x=-1\). So one point is \((-1,0)\)
Let \(y = 1\):
\(1=\log_{2}(-3x - 2)\)
Then \(-3x - 2=2^{1}=2\)
\(-3x=2 + 2=4\), \(x=-\frac{4}{3}\). So another point is \((-\frac{4}{3},1)\)

Answer:

• Asymptote point: \((-\frac{2}{3},0)\)
• Graph points: \((-1,0)\) and \((-\frac{4}{3},1)\)