Math / Algebra

Question27 28 29
Graph the logarithmic function $g(x)=\log _{3}(x+3)$ . To do this, plot two points on the graph of the function, and also draw the asymptote. Then, click on the graph-a-function button. Additionally, give the domain and range of the function using interval notation.
Domain: $\square$ (ㅁ, $\square$ ) $[\square, \square]$ ( $\square, \square$ ] $\square$ ㅁ) $\varnothing$ ロUロ
Range: $\square$ $\infty$ $-\infty$

Studdy Solution

STEP 1

What is this asking? We need to draw the graph of a logarithmic function by finding two points and the asymptote, and then state its domain and range. Watch out! Remember that logarithmic functions have a vertical asymptote where the argument becomes zero, and their domain is restricted by this asymptote.

STEP 2

1. Find the Asymptote
2. Find Two Points
3. State Domain and Range

STEP 3

The asymptote of a logarithmic function occurs when the argument of the logarithm is equal to zero.
In our case, the function is $g(x) = \log_3(x+3)$ .
So, we set the argument $x+3$ equal to zero: $x + 3 = 0$ .

STEP 4

To solve for $x$ , we subtract $3$ from both sides of the equation: $x + 3 - 3 = 0 - 3$ , which simplifies to $x = \mathbf{-3}$ .
This means the vertical asymptote is at $x = \mathbf{-3}$ .
Awesome!

STEP 5

Let's find the y-intercept, where $x = 0$ .
Plugging $x = 0$ into the function, we get $g(0) = \log_3(0+3) = \log_3(3)$ .
Since $3^1 = 3$ , we have $\log_3(3) = 1$ .
So our first point is $(0, \mathbf{1})$ .

STEP 6

Now, let's pick another $x$ value.
How about $x = 6$ ?
Plugging this in, we get $g(6) = \log_3(6+3) = \log_3(9)$ .
Since $3^2 = 9$ , we have $\log_3(9) = 2$ .
So, our second point is $(6, \mathbf{2})$ .
Perfect!

STEP 7

The domain of a logarithmic function is all the $x$ values greater than the vertical asymptote.
Since our asymptote is at $x = -3$ , the domain is $(\mathbf{-3}, \infty)$ .

STEP 8

The range of a logarithmic function is all real numbers, which we write as $(-\infty, \infty)$ .

STEP 9

Draw the vertical asymptote at $x = \mathbf{-3}$ .
Plot the points $(0, \mathbf{1})$ and $(6, \mathbf{2})$ .
Click the "graph-a-function" button.
The domain is $(\mathbf{-3}, \infty)$ and the range is $(-\infty, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the Asymptote
2. Find Two Points
3. State Domain and Range

STEP 3

The asymptote of a logarithmic function occurs when the argument of the logarithm is equal to zero.
In our case, the function is $g(x) = \log_3(x+3)$ .
So, we set the argument $x+3$ equal to zero: $x + 3 = 0$ .

STEP 4

To solve for $x$ , we subtract $3$ from both sides of the equation: $x + 3 - 3 = 0 - 3$ , which simplifies to $x = \mathbf{-3}$ .
This means the vertical asymptote is at $x = \mathbf{-3}$ .
Awesome!

STEP 5

Let's find the y-intercept, where $x = 0$ .
Plugging $x = 0$ into the function, we get $g(0) = \log_3(0+3) = \log_3(3)$ .
Since $3^1 = 3$ , we have $\log_3(3) = 1$ .
So our first point is $(0, \mathbf{1})$ .

STEP 6

Now, let's pick another $x$ value.
How about $x = 6$ ?
Plugging this in, we get $g(6) = \log_3(6+3) = \log_3(9)$ .
Since $3^2 = 9$ , we have $\log_3(9) = 2$ .
So, our second point is $(6, \mathbf{2})$ .
Perfect!

STEP 7

The domain of a logarithmic function is all the $x$ values greater than the vertical asymptote.
Since our asymptote is at $x = -3$ , the domain is $(\mathbf{-3}, \infty)$ .

STEP 8

The range of a logarithmic function is all real numbers, which we write as $(-\infty, \infty)$ .

STEP 9

Draw the vertical asymptote at $x = \mathbf{-3}$ .
Plot the points $(0, \mathbf{1})$ and $(6, \mathbf{2})$ .
Click the "graph-a-function" button.
The domain is $(\mathbf{-3}, \infty)$ and the range is $(-\infty, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the Asymptote2. Find Two Points3. State Domain and Range

STEP 3

The **asymptote** of a logarithmic function occurs when the argument of the logarithm is equal to zero.In our case, the function is g(x)=log⁡3(x+3)g(x) = \log_3(x+3)g(x)=log3​(x+3).So, we set the argument x+3x+3x+3 equal to zero: x+3=0x + 3 = 0x+3=0.

STEP 4

To solve for xxx, we subtract 333 from both sides of the equation: x+3−3=0−3x + 3 - 3 = 0 - 3x+3−3=0−3, which simplifies to x=−3x = \mathbf{-3}x=−3.This means the **vertical asymptote** is at x=−3x = \mathbf{-3}x=−3.Awesome!

STEP 5

STEP 6

Now, let's pick another xxx value.How about x=6x = 6x=6?Plugging this in, we get g(6)=log⁡3(6+3)=log⁡3(9)g(6) = \log_3(6+3) = \log_3(9)g(6)=log3​(6+3)=log3​(9).Since 32=93^2 = 932=9, we have log⁡3(9)=2\log_3(9) = 2log3​(9)=2.So, our second point is (6,2)(6, \mathbf{2})(6,2).Perfect!

STEP 7

The **domain** of a logarithmic function is all the xxx values greater than the vertical asymptote.Since our asymptote is at x=−3x = -3x=−3, the domain is (−3,∞)(\mathbf{-3}, \infty)(−3,∞).

STEP 8

The **range** of a logarithmic function is all real numbers, which we write as (−∞,∞)(-\infty, \infty)(−∞,∞).

STEP 9

Draw the vertical asymptote at x=−3x = \mathbf{-3}x=−3.Plot the points (0,1)(0, \mathbf{1})(0,1) and (6,2)(6, \mathbf{2})(6,2).Click the "graph-a-function" button.The domain is (−3,∞)(\mathbf{-3}, \infty)(−3,∞) and the range is (−∞,∞)(-\infty, \infty)(−∞,∞).

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1. Find the Asymptote
2. Find Two Points
3. State Domain and Range

The asymptote of a logarithmic function occurs when the argument of the logarithm is equal to zero.
In our case, the function is $g(x) = \log_3(x+3)$ .
So, we set the argument $x+3$ equal to zero: $x + 3 = 0$ .

To solve for $x$ , we subtract $3$ from both sides of the equation: $x + 3 - 3 = 0 - 3$ , which simplifies to $x = \mathbf{-3}$ .
This means the vertical asymptote is at $x = \mathbf{-3}$ .
Awesome!

Now, let's pick another $x$ value.
How about $x = 6$ ?
Plugging this in, we get $g(6) = \log_3(6+3) = \log_3(9)$ .
Since $3^2 = 9$ , we have $\log_3(9) = 2$ .
So, our second point is $(6, \mathbf{2})$ .
Perfect!

The domain of a logarithmic function is all the $x$ values greater than the vertical asymptote.
Since our asymptote is at $x = -3$ , the domain is $(\mathbf{-3}, \infty)$ .

The range of a logarithmic function is all real numbers, which we write as $(-\infty, \infty)$ .

Draw the vertical asymptote at $x = \mathbf{-3}$ .
Plot the points $(0, \mathbf{1})$ and $(6, \mathbf{2})$ .
Click the "graph-a-function" button.
The domain is $(\mathbf{-3}, \infty)$ and the range is $(-\infty, \infty)$ .