Question

graph the given functions, f and g, in the same rectangular coordinate system. use the integer values of x given to the right of the functions to obtain ordered pairs. describe how the graph of g is related to the graph of f.

f(x) = \sqrt{x}

g(x) = \sqrt{x + 3}

for f(x):

x = 0, 1, 4, 9

fill in the appropriate values for f(x).

(simplify your answers.)

x	f(x)
1	1
4	2
9	3

for g(x):

x = -3, -2, 1, 6

fill in the appropriate values for g(x).

(simplify your answers.)

x	g(x)
-2	1
1	\square
6	\square

Explanation:

Step1: Complete f(x) table

For \( f(x)=\sqrt{x} \), when \( x = 1 \), \( f(1)=\sqrt{1}=1 \); when \( x = 4 \), \( f(4)=\sqrt{4}=2 \); when \( x = 9 \), \( f(9)=\sqrt{9}=3 \). Wait, the existing table for \( f(x) \) has \( x = 0,1,4,9 \)? Wait, the first table for \( f(x) \) has \( x = 0,1 \) (with \( f(x)=0,1 \)) and then \( x = 4,9 \) (need to fill \( f(x) \)). Wait, \( x = 4 \): \( f(4)=\sqrt{4}=2 \), \( x = 9 \): \( f(9)=\sqrt{9}=3 \). Wait, the middle table for \( f(x) \) has \( x = 4,9 \), so \( f(4)=2 \), \( f(9)=3 \) (which matches the shaded values). Now the last table for \( f(x) \)? Wait, no, the first \( f(x) \) table: \( x = 0,1,4,9 \). Wait, the user's table for \( f(x) \) has \( x = 0,1 \) (f(x)=0,1), then \( x = 4,9 \) (f(x) to fill). Wait, \( x = 1 \): \( f(1)=\sqrt{1}=1 \) (already filled), \( x = 4 \): \( \sqrt{4}=2 \), \( x = 9 \): \( \sqrt{9}=3 \). Now for \( g(x)=\sqrt{x + 3} \), the table has \( x=-3,-2,1,6 \) (wait, the given \( x \) for \( g(x) \) is \( x=-3,-2,1,6 \)? Wait, the problem says for \( g(x) \), \( x=-3,-2,1,6 \). Let's compute \( g(x) \) for these \( x \):

• \( x=-3 \): \( g(-3)=\sqrt{-3 + 3}=\sqrt{0}=0 \)
• \( x=-2 \): \( g(-2)=\sqrt{-2 + 3}=\sqrt{1}=1 \)
• \( x=1 \): \( g(1)=\sqrt{1 + 3}=\sqrt{4}=2 \)
• \( x=6 \): \( g(6)=\sqrt{6 + 3}=\sqrt{9}=3 \)

Wait, the table for \( g(x) \) has \( x=-3,-2,1,6 \), and we need to fill \( g(x) \). Wait, the first \( g(x) \) table: \( x=-3,-2 \) (g(x)=0,1), then \( x=1,6 \) (to fill). So \( x=1 \): \( g(1)=\sqrt{1 + 3}=2 \), \( x=6 \): \( g(6)=\sqrt{6 + 3}=3 \). Now, the relationship between \( g(x) \) and \( f(x) \): \( g(x)=f(x + 3) \), so the graph of \( g(x) \) is the graph of \( f(x) \) shifted left by 3 units? Wait, no: \( f(x)=\sqrt{x} \), \( g(x)=\sqrt{x + 3}=f(x + 3) \), which is a horizontal shift left by 3 units.

But let's focus on filling the tables. For \( f(x) \) table (x=0,1,4,9):

• x=0: f(0)=√0=0
• x=1: f(1)=√1=1
• x=4: f(4)=√4=2
• x=9: f(9)=√9=3

For \( g(x) \) table (x=-3,-2,1,6):

• x=-3: g(-3)=√(-3 + 3)=√0=0
• x=-2: g(-2)=√(-2 + 3)=√1=1
• x=1: g(1)=√(1 + 3)=√4=2
• x=6: g(6)=√(6 + 3)=√9=3

Now, the last part of the \( g(x) \) table: x=1,6. So for x=1, g(1)=2; x=6, g(6)=3.

Answer:

For \( f(x) \) table (x=4,9): f(4)=2, f(9)=3.
For \( g(x) \) table (x=1,6): g(1)=2, g(6)=3.
The graph of \( g(x) \) is the graph of \( f(x) \) shifted left by 3 units (since \( g(x)=f(x + 3) \)).