Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the functions domain and range.

$y = \log_{4}(x + 3)$

$x$	$x + 3$	$y = \log_{4}(x + 3)$
$-\frac{11}{4}$
$-2$
$1$
$13$

Explanation:

Step1: Calculate \( x + 3 \) for \( x = -\frac{47}{16} \)

Substitute \( x = -\frac{47}{16} \) into \( x + 3 \).
\( x + 3=-\frac{47}{16}+3 = -\frac{47}{16}+\frac{48}{16}=\frac{1}{16} \)

Step2: Calculate \( y=\log_{4}(x + 3) \) for \( x + 3=\frac{1}{16} \)

Use the logarithm property \( \log_{a}b^{c}=c\log_{a}b \) and \( \log_{a}a = 1 \).
Since \( \frac{1}{16}=4^{-2} \), then \( y=\log_{4}(4^{-2})=- 2\log_{4}4=-2\times1=-2 \)

For \( x = -\frac{11}{4} \):

Step3: Calculate \( x + 3 \)

\( x + 3=-\frac{11}{4}+3=-\frac{11}{4}+\frac{12}{4}=\frac{1}{4} \)

Step4: Calculate \( y=\log_{4}(x + 3) \)

Since \( \frac{1}{4}=4^{-1} \), \( y=\log_{4}(4^{-1})=-1\log_{4}4=-1 \)

For \( x=-2 \):

Step5: Calculate \( x + 3 \)

\( x + 3=-2 + 3=1 \)

Step6: Calculate \( y=\log_{4}(x + 3) \)

\( y=\log_{4}1 = 0 \) (because \( \log_{a}1 = 0 \) for \( a>0,a
eq1 \))

For \( x = 1 \):

Step7: Calculate \( x + 3 \)

\( x + 3=1+3 = 4 \)

Step8: Calculate \( y=\log_{4}(x + 3) \)

\( y=\log_{4}4=1 \) (because \( \log_{a}a = 1 \))

For \( x = 13 \):

Step9: Calculate \( x + 3 \)

\( x + 3=13 + 3=16 \)

Step10: Calculate \( y=\log_{4}(x + 3) \)

Since \( 16 = 4^{2} \), \( y=\log_{4}(4^{2})=2\log_{4}4 = 2 \)

Now, fill the table:

\( x \)	\( x + 3 \)	\( y=\log_{4}(x + 3) \)
\( -\frac{11}{4} \)	\( \frac{1}{4} \)	\( -1 \)
\( -2 \)	\( 1 \)	\( 0 \)
\( 1 \)	\( 4 \)	\( 1 \)
\( 13 \)	\( 16 \)	\( 2 \)

To find the domain: The argument of the logarithm \( x + 3>0\Rightarrow x>-3 \), so domain is \( (-3,\infty) \)

To find the range: The range of a logarithmic function \( \log_{a}(u(x)) \) where \( u(x) \) is a linear function (here \( u(x)=x + 3 \)) with \( u(x)>0 \) is \( (-\infty,\infty) \), so range is \( (-\infty,\infty) \)

Answer:

Table Filling:

• For \( x = -\frac{47}{16} \): \( x + 3=\frac{1}{16} \), \( y=-2 \)
• For \( x = -\frac{11}{4} \): \( x + 3=\frac{1}{4} \), \( y=-1 \)
• For \( x=-2 \): \( x + 3 = 1 \), \( y = 0 \)
• For \( x = 1 \): \( x + 3=4 \), \( y = 1 \)
• For \( x = 13 \): \( x + 3=16 \), \( y = 2 \)

Domain and Range:

• Domain: \( (-3,\infty) \)
• Range: \( (-\infty,\infty) \)