Graph the logarithm. Make sure to list the domain and to include any vertical asymptotes on the graph. $(x)=-\log _{3}(x+4)+2$

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Q. Graph the logarithm. Make sure to list the domain and to include any vertical asymptotes on the graph.

(x)=-\log _{3}(x+4)+2

Understand Logarithmic Function Shape: To graph the function $f(x) = -\log_{3}(x+4) + 2$ , we first need to understand the basic shape of a logarithmic function. The parent function $\log_{b}(x)$ has a vertical asymptote at $x = 0$ , passes through the point ( $1$ , $0$ ), and is increasing if $b > 1$ . Since our function has a negative sign in front of the logarithm and a vertical shift upwards by $2$ units, we expect the graph to be reflected over the x-axis and shifted up.
Determine Domain: Next, we need to determine the domain of the function. The argument of the logarithm, $x+4$ , must be greater than zero because the logarithm of a non-positive number is undefined. Therefore, we set up the inequality $x+4 > 0$ and solve for $x$ . $\newline$ $x+4 > 0$ $\newline$ $x > -4$ $\newline$ This tells us that the domain of the function is $x > -4$ .
Vertical Asymptote: The vertical asymptote of the function will be at the value of $x$ that makes the argument of the logarithm zero, which is $x = -4$ . This is because as $x$ approaches $-4$ from the right, the logarithm will approach negative infinity, and the negative sign in front of the logarithm will flip it to positive infinity, but after adding $2$ , it will still be positive infinity.
Sketch Graph: Now we can sketch the graph. We start by drawing a vertical dashed line at $x = -4$ to represent the vertical asymptote. Then, we plot a few key points to help us draw the curve. We know that the logarithm function has a key point at ( $1$ , $0$ ), but we need to adjust this for our function. We set the inside of the logarithm equal to $1$ and solve for $x$ : $\newline$ $x+4 = 1$ $\newline$ $x = -3$ $\newline$ So, the corresponding point on our graph will be $(-3, 2)$ because we need to reflect it over the x-axis and shift it up by $2$ units.
Find Key Points: We can also find another point by setting the inside of the logarithm equal to $3$ (since the base is $3$ ) and solving for $x$ : $\newline$ $x+4 = 3$ $\newline$ $x = -1$ $\newline$ The corresponding y-value when we plug $x = -1$ into the function is: $\newline$ $f(-1) = -\log_{3}(-1+4) + 2 = -\log_{3}(3) + 2 = -1 + 2 = 1$ $\newline$ So another point on our graph is $(-1, 1)$ .
Draw Curve: With these points and the vertical asymptote, we can now sketch the graph. It will be a curve that approaches the vertical asymptote at $x = -4$ and passes through the points $(-3, 2)$ and $(-1, 1)$ , decreasing as $x$ increases because of the negative sign in front of the logarithm.