Math / Trigonometry

QuestionSketch the graph of the function. $f(x)=\arctan (4 x)$
Compare the graph to the graph of the parent inverse trigonometric function. The graph of $f(x)$ is $\arctan (x)$ with a ---Select--- $\square$ Read It

Studdy Solution

STEP 1

What is this asking? We need to graph $f(x) = \arctan(4x)$ and explain how it's different from the graph of $\arctan(x)$ . Watch out! Don't mix up $\arctan(x)$ with $\tan(x)$ !
Also, remember $\arctan(x)$ has horizontal asymptotes, while $\tan(x)$ has vertical asymptotes.

STEP 2

1. Graph the parent function
2. Graph our function
3. Compare the graphs

STEP 3

The graph of $\arctan(x)$ has horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ .
It passes through the origin $(0, 0)$ .
It increases from left to right, taking on values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ .

STEP 4

Draw the horizontal asymptotes as dashed lines.
Mark the origin.
Sketch a smooth curve increasing between the asymptotes, passing through the origin.

STEP 5

$f(x) = \arctan(4x)$ will still have horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ .
This is because the $4x$ inside the $\arctan$ affects the horizontal stretch of the graph, but not the vertical asymptotes.

STEP 6

Let's find a nice point to help us graph $f(x)$ .
When $4x = 1$ , we have $f(x) = \arctan(1) = \frac{\pi}{4}$ .
So, when $x = \frac{1}{4}$ , $f(x) = \frac{\pi}{4}$ .
The point $(\frac{1}{4}, \frac{\pi}{4})$ is on our graph!

STEP 7

Draw the horizontal asymptotes (same as before).
Plot the point $(\frac{1}{4}, \frac{\pi}{4})$ .
Sketch a smooth curve increasing between the asymptotes, passing through this new point.

STEP 8

Notice that $f(x) = \arctan(4x)$ is a horizontal compression of $\arctan(x)$ by a factor of $\frac{1}{4}$ .
This means the graph of $f(x)$ is "squished" horizontally towards the $y$ -axis compared to $\arctan(x)$ .
It looks steeper!

STEP 9

The graph of $f(x) = \arctan(4x)$ is a horizontal compression of $\arctan(x)$ by a factor of $\frac{1}{4}$ .

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QuestionSketch the graph of the function. $f(x)=\arctan (4 x)$
Compare the graph to the graph of the parent inverse trigonometric function. The graph of $f(x)$ is $\arctan (x)$ with a ---Select--- $\square$ Read It

Studdy Solution

STEP 1

What is this asking? We need to graph $f(x) = \arctan(4x)$ and explain how it's different from the graph of $\arctan(x)$ . Watch out! Don't mix up $\arctan(x)$ with $\tan(x)$ !
Also, remember $\arctan(x)$ has horizontal asymptotes, while $\tan(x)$ has vertical asymptotes.

STEP 2

1. Graph the parent function
2. Graph our function
3. Compare the graphs

STEP 3

The graph of $\arctan(x)$ has horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ .
It passes through the origin $(0, 0)$ .
It increases from left to right, taking on values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ .

STEP 4

Draw the horizontal asymptotes as dashed lines.
Mark the origin.
Sketch a smooth curve increasing between the asymptotes, passing through the origin.

STEP 5

$f(x) = \arctan(4x)$ will still have horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ .
This is because the $4x$ inside the $\arctan$ affects the horizontal stretch of the graph, but not the vertical asymptotes.

STEP 6

Let's find a nice point to help us graph $f(x)$ .
When $4x = 1$ , we have $f(x) = \arctan(1) = \frac{\pi}{4}$ .
So, when $x = \frac{1}{4}$ , $f(x) = \frac{\pi}{4}$ .
The point $(\frac{1}{4}, \frac{\pi}{4})$ is on our graph!

STEP 7

Draw the horizontal asymptotes (same as before).
Plot the point $(\frac{1}{4}, \frac{\pi}{4})$ .
Sketch a smooth curve increasing between the asymptotes, passing through this new point.

STEP 8

Notice that $f(x) = \arctan(4x)$ is a horizontal compression of $\arctan(x)$ by a factor of $\frac{1}{4}$ .
This means the graph of $f(x)$ is "squished" horizontally towards the $y$ -axis compared to $\arctan(x)$ .
It looks steeper!

STEP 9

The graph of $f(x) = \arctan(4x)$ is a horizontal compression of $\arctan(x)$ by a factor of $\frac{1}{4}$ .

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QuestionSketch the graph of the function. f(x)=arctan⁡(4x)f(x)=\arctan (4 x)f(x)=arctan(4x)Compare the graph to the graph of the parent inverse trigonometric function. The graph of f(x)f(x)f(x) is arctan⁡(x)\arctan (x)arctan(x) with a ---Select--- □\square□ Read It

Studdy Solution

STEP 1

STEP 2

1. Graph the parent function2. Graph our function3. Compare the graphs

STEP 3

STEP 4

Draw the horizontal asymptotes as dashed lines.Mark the origin.Sketch a smooth curve increasing between the asymptotes, passing through the origin.

STEP 5

STEP 6

STEP 7

Draw the horizontal asymptotes (same as before).Plot the point (14,π4)(\frac{1}{4}, \frac{\pi}{4})(41​,4π​).Sketch a smooth curve increasing between the asymptotes, passing through this new point.

STEP 8

STEP 9

The graph of f(x)=arctan⁡(4x)f(x) = \arctan(4x)f(x)=arctan(4x) is a horizontal compression of arctan⁡(x)\arctan(x)arctan(x) by a factor of 14\frac{1}{4}41​.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionSketch the graph of the function. $f(x)=\arctan (4 x)$
Compare the graph to the graph of the parent inverse trigonometric function. The graph of $f(x)$ is $\arctan (x)$ with a ---Select--- $\square$ Read It

1. Graph the parent function
2. Graph our function
3. Compare the graphs

Draw the horizontal asymptotes as dashed lines.
Mark the origin.
Sketch a smooth curve increasing between the asymptotes, passing through the origin.

Draw the horizontal asymptotes (same as before).
Plot the point $(\frac{1}{4}, \frac{\pi}{4})$ .
Sketch a smooth curve increasing between the asymptotes, passing through this new point.

The graph of $f(x) = \arctan(4x)$ is a horizontal compression of $\arctan(x)$ by a factor of $\frac{1}{4}$ .