Answer:
a)
b)
[tex]Domain:(-4,\infty)[/tex]
[tex]Range:(-\infty,\infty)[/tex]
c) x-intercept is 0
Explanation:
Given:
[tex]f(x)=\log_2(x+4)-2[/tex]
a) See below the graph of different transformations of the given function;
The equation of the vertical asymptote as shown on the graph is x = -4
b) The domain of a function is the set of possible input values for which the function is defined. The domain of a graph is the set of possible values from left to right.
Looking at the given graph, we can see that the domain of the function is;
[tex]Domain:(-4,\infty)[/tex]
The range of a graph is the set of values from the bottom to the top of the graph. Looking at the graph, we can see that the range is;
[tex]Range:(-\infty,\infty)[/tex]
c) We'll follow the below steps to determine the x-intercept of the function;
Step 1: Substitute f(x) with 0;
[tex]0=\log_2(x+4)-2[/tex]
Step 2: Add 2 to both sides;
[tex]\begin{gathered} 0+2=\log_2(x+4)-2+2 \\ 2=\log_2(x+4) \end{gathered}[/tex]
Step 3: Apply the below rule;
[tex]\begin{gathered} \log_ab=c \\ b=a^c \end{gathered}[/tex][tex]\begin{gathered} 2^2=x+4 \\ 4=x+4 \\ 4-4=x \\ 0=x \\ \therefore x=0 \end{gathered}[/tex]
So the x-intercept is 0
