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 \( x + 3 \) when \( x = -\frac{47}{16} \)

We substitute \( x = -\frac{47}{16} \) into \( x + 3 \). First, rewrite \( 3 \) as \( \frac{48}{16} \), then \( x + 3 = -\frac{47}{16}+\frac{48}{16}=\frac{1}{16} \).

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

We know that \( \log_{a}b = c \) is equivalent to \( a^{c}=b \). Let \( \log_{4}\frac{1}{16}=y \), then \( 4^{y}=\frac{1}{16} \). Since \( 4 = 2^{2} \) and \( \frac{1}{16}=2^{-4}=(2^{2})^{-2}=4^{-2} \), so \( 4^{y}=4^{-2} \), which means \( y = - 2 \).

For the other values (we'll do one more as an example, say \( x=-2 \)):

Step3: Calculate \( x + 3 \) when \( x=-2 \)

Substitute \( x = - 2 \) into \( x + 3 \), we get \( -2 + 3=1 \).

Step4: Calculate \( y=\log_{4}(x + 3) \) when \( x + 3 = 1 \)

Using the property \( \log_{a}1 = 0 \) (since \( a^{0}=1 \) for \( a>0,a
eq1 \)), so \( \log_{4}1 = 0 \).

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

Step5: Calculate \( x + 3 \) when \( x = -\frac{11}{4} \)

Rewrite \( 3 \) as \( \frac{12}{4} \), then \( x + 3=-\frac{11}{4}+\frac{12}{4}=\frac{1}{4} \).

Step6: Calculate \( y=\log_{4}(x + 3) \) when \( x + 3=\frac{1}{4} \)

Let \( \log_{4}\frac{1}{4}=y \), then \( 4^{y}=\frac{1}{4}=4^{-1} \), so \( y=-1 \).

For \( x = 1 \):

Step7: Calculate \( x + 3 \) when \( x = 1 \)

Substitute \( x = 1 \) into \( x + 3 \), we get \( 1+3 = 4 \).

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

Using the property \( \log_{a}a = 1 \), so \( \log_{4}4 = 1 \).

For \( x = 13 \):

Step9: Calculate \( x + 3 \) when \( x = 13 \)

Substitute \( x = 13 \) into \( x + 3 \), we get \( 13 + 3=16 \).

Step10: Calculate \( y=\log_{4}(x + 3) \) when \( x + 3 = 16 \)

Let \( \log_{4}16=y \), then \( 4^{y}=16 \). Since \( 16 = 4^{2} \), so \( y = 2 \).

Now, filling 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 \( x + 3>0 \), so \( x>-3 \), so the domain is \( (-3,\infty) \).

To find the range: The range of a logarithmic function \( \log_{a}(x + h)+k \) (here \( h = 3,k = 0,a = 4 \)) is \( (-\infty,\infty) \) (all real numbers), so the range is \( (-\infty,\infty) \) or \( \mathbb{R} \).

Answer:

For the table (first row filled):

• When \( x = -\frac{47}{16} \), \( x + 3=\boldsymbol{\frac{1}{16}} \), \( y=\log_{4}(x + 3)=\boldsymbol{-2} \)

Domain: \( \boldsymbol{(-3,\infty)} \)

Range: \( \boldsymbol{(-\infty,\infty)} \) (or \( \boldsymbol{\mathbb{R}} \))