Answer:
Correct answer option A.
Step-by-step explanation:
We need to figure out the graph of
[tex]y= log_3(x + 2)[/tex]
First of all, let us discuss graph of:
[tex]y=log_3x[/tex]
[tex]y=log_31 = 0[/tex] ([tex]\because[/tex] log of 1 is always 0 irrespective of the base)
i.e. point (1, 0) lies on the graph.
- And when we put x = 1, the log function tends to -[tex]\infty[/tex].
i.e. [tex]x\rightarrow 1 \Rightarrow y\rightarrow -\infty[/tex]
[tex]y =log_33 = 1[/tex] i.e. point (3, 1) lies on the graph
Graph of logarithmic function is always increasing.
Now, let us consider the graph of [tex]y= log_3(x + 2)[/tex]
- Putting x = -1, [tex]y= log_3(-1 + 2) = log_31 = 0[/tex] So, the point that lies on graph is (-1, 0)
- Putting x = 1, [tex]y= log_3(1 + 2) = log_33 = 1[/tex] So, the point that lies on graph is (1,1).
- Putting x = -2, [tex]y= log_3(-2 + 2) = log_30 \rightarrow -\infty[/tex]
Please refer to attached graph as well.
Correct answer option A.
