Solved on Mar 27, 2024

Find the equation of the function $f(x)$ where $f(x)=\frac{1}{x+3}$ .

STEP 1

Assumptions
1. We are working with the function $f(x) = \frac{1}{x+3}$ .
2. We want to graph this function.
3. We want to identify key features of the graph such as intercepts, asymptotes, and intervals of increase or decrease.

STEP 1

Assumptions
1. The function given is $f(x) = \frac{1}{x+3}$ .
2. To graph the function, we need to understand its behavior and identify key features such as intercepts, asymptotes, and intervals of increase or decrease.
3. To "solve the function" typically means finding the values of $x$ for which $f(x)$ is defined or finding the range of the function. Since $f(x)$ is already solved for $y$ , we will interpret "solve the function" as finding the domain and range of $f(x)$ .

STEP 2

To graph the function, we first identify any vertical asymptotes. A vertical asymptote occurs where the function is undefined, which is when the denominator is zero.
$x + 3 = 0$

STEP 2

First, let's identify the vertical asymptote of the function. A vertical asymptote occurs where the function is undefined, which is when the denominator of the function equals zero.
$x + 3 = 0$

STEP 3

Solve for $x$ to find the vertical asymptote.
$x = -3$

STEP 3

Solve for $x$ to find the vertical asymptote.
$x = -3$

STEP 4

The vertical asymptote is at $x = -3$ . This means the graph of the function will approach this line but never touch or cross it.

STEP 4

Next, we identify any horizontal asymptotes. For rational functions, if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$ .

STEP 5

Since the numerator is a constant (1) and the denominator is a linear term ( $x + 3$ ), the degree of the numerator is 0 and the degree of the denominator is 1. Therefore, there is a horizontal asymptote at $y = 0$ .

STEP 5

Next, we will determine the horizontal asymptote. For rational functions where the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the x-axis, or $y = 0$ .

STEP 6

Now, we find the $y$ -intercept by setting $x = 0$ and solving for $f(x)$ .
$f(0) = \frac{1}{0+3}$

STEP 6

There are no x-intercepts for this function because the numerator, 1, is never zero.

STEP 7

There is a y-intercept when $x = 0$ . We can find this by evaluating $f(0)$ .
$f(0) = \frac{1}{0+3}$

STEP 7

Calculate the $y$ -intercept.
$f(0) = \frac{1}{3}$

STEP 8

Calculate the y-intercept.
$f(0) = \frac{1}{3}$

STEP 8

The $y$ -intercept is the point $(0, \frac{1}{3})$ .

STEP 9

Since there is no $x$ -intercept (the function never crosses the $x$ -axis), we can now plot the vertical asymptote, the horizontal asymptote, and the $y$ -intercept on a graph.

STEP 9

Now we have enough information to sketch the graph of the function. The graph will have a vertical asymptote at $x = -3$ , a horizontal asymptote at $y = 0$ , and will pass through the point (0, 1/3).

STEP 10

To get a better idea of the shape of the graph, we can choose a few $x$ -values on either side of the vertical asymptote and calculate the corresponding $y$ -values.

STEP 10

To solve for the domain of the function, we look for all values of $x$ for which the function is defined. Since the only restriction is the vertical asymptote at $x = -3$ , the domain is all real numbers except $x = -3$ .
$Domain: x \in \mathbb{R}, x \neq -3$

STEP 11

To solve for the range of the function, we look for all possible values of $f(x)$ . Since $f(x)$ can approach any real number except 0 (due to the horizontal asymptote), the range is all real numbers except $y = 0$ .
$Range: y \in \mathbb{R}, y \neq 0$
The graph of the function $f(x) = \frac{1}{x+3}$ has a vertical asymptote at $x = -3$ , a horizontal asymptote at $y = 0$ , and passes through the point (0, 1/3). The domain of the function is all real numbers except $x = -3$ , and the range is all real numbers except $y = 0$ .

STEP 11

Choose $x = -4$ (to the left of the vertical asymptote) and calculate $f(-4)$ .
$f(-4) = \frac{1}{-4+3} = \frac{1}{-1} = -1$

STEP 12

Choose $x = -2$ (to the right of the vertical asymptote) and calculate $f(-2)$ .
$f(-2) = \frac{1}{-2+3} = \frac{1}{1} = 1$

STEP 13

Plot the points $(-4, -1)$ and $(-2, 1)$ on the graph.

STEP 14

To understand the behavior near the asymptotes, consider the following:
- As $x$ approaches -3 from the left, $f(x)$ decreases without bound (goes to negative infinity). - As $x$ approaches -3 from the right, $f(x)$ increases without bound (goes to positive infinity). - As $x$ goes to positive or negative infinity, $f(x)$ approaches the horizontal asymptote $y = 0$ .

STEP 15

Sketch the curve using the points and asymptotic behavior. The graph will approach the asymptotes but never touch or cross them.

STEP 16

Check for symmetry. The function $f(x) = \frac{1}{x+3}$ is not symmetric with respect to the $y$ -axis (not even), the $x$ -axis (not odd), or the origin.

STEP 17

The graph of $f(x) = \frac{1}{x+3}$ is complete. It should show a curve approaching the vertical asymptote $x = -3$ and the horizontal asymptote $y = 0$ , with a $y$ -intercept at $(0, \frac{1}{3})$ .
To solve the function $f(x) = \frac{1}{x+3}$ for a given $x$ , simply substitute the value of $x$ into the function and calculate $f(x)$ .

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Find the equation of the function f(x)f(x)f(x) where f(x)=1x+3f(x)=\frac{1}{x+3}f(x)=x+31​.

STEP 1

Assumptions1. We are working with the function f(x)=1x+3 f(x) = \frac{1}{x+3} f(x)=x+31​.2. We want to graph this function.3. We want to identify key features of the graph such as intercepts, asymptotes, and intervals of increase or decrease.

STEP 1

STEP 2

To graph the function, we first identify any vertical asymptotes. A vertical asymptote occurs where the function is undefined, which is when the denominator is zero.x+3=0x + 3 = 0x+3=0

STEP 2

First, let's identify the vertical asymptote of the function. A vertical asymptote occurs where the function is undefined, which is when the denominator of the function equals zero.x+3=0x + 3 = 0x+3=0

STEP 3

Solve for x x x to find the vertical asymptote.x=−3x = -3x=−3

STEP 3

Solve for x x x to find the vertical asymptote.x=−3x = -3x=−3

STEP 4

The vertical asymptote is at x=−3 x = -3 x=−3. This means the graph of the function will approach this line but never touch or cross it.

STEP 4

Next, we identify any horizontal asymptotes. For rational functions, if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0 y = 0 y=0.

STEP 5

Since the numerator is a constant (1) and the denominator is a linear term (x+3 x + 3 x+3), the degree of the numerator is 0 and the degree of the denominator is 1. Therefore, there is a horizontal asymptote at y=0 y = 0 y=0.

STEP 5

Next, we will determine the horizontal asymptote. For rational functions where the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the x-axis, or y=0 y = 0 y=0.

STEP 6

Now, we find the y y y-intercept by setting x=0 x = 0 x=0 and solving for f(x) f(x) f(x).f(0)=10+3f(0) = \frac{1}{0+3}f(0)=0+31​

STEP 6

There are no x-intercepts for this function because the numerator, 1, is never zero.

STEP 7

There is a y-intercept when x=0 x = 0 x=0. We can find this by evaluating f(0) f(0) f(0).f(0)=10+3f(0) = \frac{1}{0+3}f(0)=0+31​

STEP 7

Calculate the y y y-intercept.f(0)=13f(0) = \frac{1}{3}f(0)=31​

STEP 8

Calculate the y-intercept.f(0)=13f(0) = \frac{1}{3}f(0)=31​

STEP 8

The y y y-intercept is the point (0,13) (0, \frac{1}{3}) (0,31​).

STEP 9

Since there is no x x x-intercept (the function never crosses the x x x-axis), we can now plot the vertical asymptote, the horizontal asymptote, and the y y y-intercept on a graph.

STEP 9

Now we have enough information to sketch the graph of the function. The graph will have a vertical asymptote at x=−3 x = -3 x=−3, a horizontal asymptote at y=0 y = 0 y=0, and will pass through the point (0, 1/3).

STEP 10

To get a better idea of the shape of the graph, we can choose a few x x x-values on either side of the vertical asymptote and calculate the corresponding y y y-values.

STEP 10

STEP 11

STEP 11

Choose x=−4 x = -4 x=−4 (to the left of the vertical asymptote) and calculate f(−4) f(-4) f(−4).f(−4)=1−4+3=1−1=−1f(-4) = \frac{1}{-4+3} = \frac{1}{-1} = -1f(−4)=−4+31​=−11​=−1

STEP 12

Choose x=−2 x = -2 x=−2 (to the right of the vertical asymptote) and calculate f(−2) f(-2) f(−2).f(−2)=1−2+3=11=1f(-2) = \frac{1}{-2+3} = \frac{1}{1} = 1f(−2)=−2+31​=11​=1

STEP 13

Plot the points (−4,−1) (-4, -1) (−4,−1) and (−2,1) (-2, 1) (−2,1) on the graph.

STEP 14

STEP 15

Sketch the curve using the points and asymptotic behavior. The graph will approach the asymptotes but never touch or cross them.

STEP 16

Check for symmetry. The function f(x)=1x+3 f(x) = \frac{1}{x+3} f(x)=x+31​ is not symmetric with respect to the y y y-axis (not even), the x x x-axis (not odd), or the origin.

STEP 17

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Find the equation of the function $f(x)$ where $f(x)=\frac{1}{x+3}$ .

Assumptions
1. We are working with the function $f(x) = \frac{1}{x+3}$ .
2. We want to graph this function.
3. We want to identify key features of the graph such as intercepts, asymptotes, and intervals of increase or decrease.

To graph the function, we first identify any vertical asymptotes. A vertical asymptote occurs where the function is undefined, which is when the denominator is zero.
$x + 3 = 0$

First, let's identify the vertical asymptote of the function. A vertical asymptote occurs where the function is undefined, which is when the denominator of the function equals zero.
$x + 3 = 0$

Solve for $x$ to find the vertical asymptote.
$x = -3$

Solve for $x$ to find the vertical asymptote.
$x = -3$

The vertical asymptote is at $x = -3$ . This means the graph of the function will approach this line but never touch or cross it.

Next, we identify any horizontal asymptotes. For rational functions, if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$ .

Since the numerator is a constant (1) and the denominator is a linear term ( $x + 3$ ), the degree of the numerator is 0 and the degree of the denominator is 1. Therefore, there is a horizontal asymptote at $y = 0$ .

Next, we will determine the horizontal asymptote. For rational functions where the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the x-axis, or $y = 0$ .

Now, we find the $y$ -intercept by setting $x = 0$ and solving for $f(x)$ .
$f(0) = \frac{1}{0+3}$

There is a y-intercept when $x = 0$ . We can find this by evaluating $f(0)$ .
$f(0) = \frac{1}{0+3}$

Calculate the $y$ -intercept.
$f(0) = \frac{1}{3}$

Calculate the y-intercept.
$f(0) = \frac{1}{3}$

The $y$ -intercept is the point $(0, \frac{1}{3})$ .

Since there is no $x$ -intercept (the function never crosses the $x$ -axis), we can now plot the vertical asymptote, the horizontal asymptote, and the $y$ -intercept on a graph.

Now we have enough information to sketch the graph of the function. The graph will have a vertical asymptote at $x = -3$ , a horizontal asymptote at $y = 0$ , and will pass through the point (0, 1/3).

To get a better idea of the shape of the graph, we can choose a few $x$ -values on either side of the vertical asymptote and calculate the corresponding $y$ -values.

Choose $x = -4$ (to the left of the vertical asymptote) and calculate $f(-4)$ .
$f(-4) = \frac{1}{-4+3} = \frac{1}{-1} = -1$

Choose $x = -2$ (to the right of the vertical asymptote) and calculate $f(-2)$ .
$f(-2) = \frac{1}{-2+3} = \frac{1}{1} = 1$

Plot the points $(-4, -1)$ and $(-2, 1)$ on the graph.

Check for symmetry. The function $f(x) = \frac{1}{x+3}$ is not symmetric with respect to the $y$ -axis (not even), the $x$ -axis (not odd), or the origin.