Answer:
For f(x);
The domain is;
[tex]D\colon x=(-\infty,\infty)[/tex]
The range is;
[tex]R\colon y=\lbrack0,\infty)[/tex]
Graphing those points for function A, we have;
The domain and range of the given function A is;
[tex]\begin{gathered} \text{Domain}\colon x=(-\infty,\infty) \\ \text{Range}\colon y=\lbrack-3,\infty) \end{gathered}[/tex]
Graphing those points for function B, we have;
The domain and range of the given function B is;
[tex]\begin{gathered} \text{Domain}\colon x=(-\infty,\infty) \\ \text{Range}\colon y=(-\infty,0\rbrack \end{gathered}[/tex]
Explanation:
Given the function in the attached image;
The function is a square function and can be written as;
[tex]f(x)=x^2[/tex]
The domain is;
[tex]D\colon x=(-\infty,\infty)[/tex]
The range is;
[tex]R\colon y=\lbrack0,\infty)[/tex]
A.
[tex]f(x-1)-3=(x-1)^2-3[/tex]
B.
[tex]-f(x)=-x^2[/tex]
Graphing the functions;
For A;
[tex]\begin{gathered} f(1-1)=(1-1)^2-3=-3 \\ (1,-3) \\ f(3-1)=(3-1)^2-3=1 \\ (3,1) \\ f(-1-1)=(-1-1)^2-3=1 \\ (-1,1) \end{gathered}[/tex]
Graphing those points for function A, we have;
The domain and range of the given function A is;
[tex]\begin{gathered} \text{Domain}\colon x=(-\infty,\infty) \\ \text{Range}\colon y=\lbrack-3,\infty) \end{gathered}[/tex]
For B;
[tex]\begin{gathered} -f(x)=-x^2 \\ -f(0)=-0^2 \\ (0,0) \\ -f(2)=-2^2=-4 \\ (2,-4) \\ -f(-2)=-(-2)^2=-4 \\ (-2,-4) \end{gathered}[/tex]
Graphing those points for function B, we have;
The domain and range of the given function B is;
[tex]\begin{gathered} \text{Domain}\colon x=(-\infty,\infty) \\ \text{Range}\colon y=(-\infty,0\rbrack \end{gathered}[/tex]
