use the function below to answer the following questions.\n\n$y = \\log_3(x + 3)$\n\n(a) use transformations…

Explanation:

Step1: Analyze the transformation

The function (y = \log_{3}(x + 3)) is a horizontal shift of (y=\log_{3}x). The transformation rule for (y = f(x + h)) is a shift of (y = f(x)) to the left by (h) units. Here (h = 3), so the graph of (y=\log_{3}x) is shifted 3 units to the left.

Step2: Find the domain

For the function (y=\log_{3}(x + 3)), the argument of the logarithm must be positive. So (x+3>0), which gives (x>- 3). In interval notation, the domain is ((-3,\infty)).

Step3: Find the range

The range of any logarithmic function of the form (y = \log_{a}(x + h)+k) (where (a>0,a\neq1)) is ((-\infty,\infty)). So for (y=\log_{3}(x + 3)), the range is ((-\infty,\infty))

Step4: Find the vertical asymptote

For the function (y=\log_{3}(x + 3)), as (x+3\rightarrow0^{+}) (i.e., (x\rightarrow - 3^{+})), (y\rightarrow-\infty). The vertical asymptote is (x=-3)

Answer:

• (a) Shift the graph of (y = \log_{3}x) 3 units to the left.
• (b) Domain: ((-3,\infty)), Range: ((-\infty,\infty))
• (c) (x=-3)