a. The given function is:
[tex]f(x)=\sqrt[]{-(x+3)}-2[/tex]
It has a restriction in the domain given that the radicand of the square root can't be negative. Then:
[tex]\begin{gathered} -(x+3)\ge0 \\ x+3\le0 \\ x\le-3 \end{gathered}[/tex]
Then, we can start with x=-3:
[tex]\begin{gathered} f(-3)=\sqrt[]{-(-3+3)}-2 \\ f(-3)=\sqrt[]{-(0)}-2 \\ f(-3)=\sqrt[]{0}-2 \\ f(-3)=-2 \end{gathered}[/tex]
The first point is (-3,-2)
Now, x=-4:
[tex]\begin{gathered} f(-4)=\sqrt[]{-(-4+3)}-2 \\ f(-4)=\sqrt[]{-(-1)}-2 \\ f(-4)=\sqrt[]{1}-2 \\ f(-4)=1-2 \\ f(-4)=-1 \end{gathered}[/tex]
The second point is (-4,-1)
And now we can evaluate f(-7):
[tex]\begin{gathered} f(-7)=\sqrt[]{-(-7+3)}-2 \\ f(-7)=\sqrt[]{-(-4)}-2 \\ f(-7)=\sqrt[]{4}-2 \\ f(-7)=2-2 \\ f(-7)=0 \end{gathered}[/tex]
Another point in the graph is (-7,0).
Let's evaluate a last point f(-10):
[tex]\begin{gathered} f(-10)=\sqrt[]{-(-10+3)}-2 \\ f(-10)=\sqrt[]{-(-7)}-2 \\ f(-10)=\sqrt[]{7}-2 \\ f(-10)=2.646-2 \\ f(-10)=0.646 \end{gathered}[/tex]
Now, the sketch of the graph will be:
b. The function is:
[tex]g(x)=-(x-1)^2+5[/tex]
This has no restrictions in the domain.
Let's evaluate g(0):
[tex]\begin{gathered} g(0)=-(0-1)^2+5_{}_{} \\ g(0)=-(-1)^2+5_{} \\ g(0)=-1+5_{} \\ g(0)=4 \end{gathered}[/tex]
The first point is (0,4).
Now, g(1):
[tex]\begin{gathered} g(1)=-(1-1)^2+5_{} \\ g(1)=-(0)^2+5_{} \\ g(1)=-0+5_{} \\ g(1)=5_{} \end{gathered}[/tex]
The second point is (1,5).
Evaluate g(3):
[tex]\begin{gathered} g(3)=-(3-1)^2+5_{} \\ g(3)=-(2)^2+5_{} \\ g(3)=-4+5_{} \\ g(3)=1 \end{gathered}[/tex]
The third point is (3,1).
And evaluate g(-1):
[tex]\begin{gathered} g(-1)=-(-1-1)^2+5_{} \\ g(-1)=-(-2)^2+5_{} \\ g(-1)=-4+5_{} \\ g(-1)=1 \end{gathered}[/tex]
Another point is (-1,1).
The sketch of the graph will look like this:
