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JulijaS [17]
1 year ago
15

Choose the graph of y = 2 tan x.

Mathematics
2 answers:
JulsSmile [24]1 year ago
7 0

Find the asymptotes

For any y=tan(x), vertical asymptotes occur at x = \frac{\pi }{2} + n\pi, where n is an integer.

Use the basic period for y=tan(x), (-\frac{\pi }2} , \frac{\pi }{2} ), to find the vertial asymptotes for y= 2 tan (x). Set the inside of the tangent function, bx+c, for y = a tan (bx+c) + d equal to -\frac{\pi }{2} to find where the vertical asymptote occurs for y=2tan(x).

x=-\frac{\pi }{2}

Set the inside of the tangent function x equal to \frac{\pi }{2}

x=\pi /2

the basic period for y=2tan(x) will occur at (-\pi/2 ,\pi /2), where -pi/2 and pi/2 are vertical asymptotes.

(-\frac{\pi }{2} , \frac{\pi }{2} )

Find the period \frac{\pi }{|b|} to find where the vertical asymptotes exist.

The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.

\frac{\pi }{1}

Divide \pi by 1

\pi

The vertical asymptotes for y = 2 tan (x)  occur at -\frac{\pi }{2} , \frac{\pi }{2,} and every \pi n, and where n is a integer.

\pi n

There are only vertical asymptotes for tangent and contagent functions.

Vertical asymptotes : x=\frac{\pi }{2} +\pi n  for any integer n.

No horizontal asymptotes

No oblique asymptotes

use a form for a tan (bx-c)+d to find variables used to find the amptitude , period, phase shift, and vertical shift.

a=2

b=1

c=0

d=0

Since the graph of the function tan does not have a maximum or minimum value, there can be no value for amplitude.

Amplitude : None

Find the period of 2 tan (x)

The period of the function can be caculated using \frac{\pi }{|b|}

\pi /|b|

Replace b with 1 in the formula for period.

\frac{\pi }{|1|}

The absolute value is the distance between a number and zero . The distance between 0 and 1 is 1.

\frac{\pi }{1}

Divide \pi by 1

\pi

Find the phase shift using the formula \frac{c}{b}

Phase Shift: \frac{c}{b}

Replace the values of c and b in the equation for phase shift

Phase shift: 0/1

Divide 0 by 1

Phase Shift; 0

Find the vertical shift d.

vertical shift: 0

List the properties of the trigonometric function

Amplitude; none

period: \pi

Phase shift: 0 ( 0 to right )

Vertical shift: 0

The trig function can be graphed using the amplitude , period, phase shift, vertical shift, and the points.

Vertical asymptotes : x= pi/2+pi n for any integer n.

Amplitude: None

Period: \pi

Phase Shift: 0 ( 0 to the right)

Vertical Shift: 0

Sindrei [870]1 year ago
3 0

Answer:

The image shows the graph of y = 2 tan x.

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tatyana61 [14]

Answer:

2.\ f(x) = \frac{3}{4}x^2 + 2x - 5

Step-by-step explanation:

Given

f(x) = -8x^3 - 16x^2 - 4x\\f(x) = \frac{3}{4}x^2 + 2x - 5\\f(x) = \frac{4}{x^2} - \frac{2}{x} + 1\\f(x) = 0x^2 - 9x + 7

Required

Which of the above is a quadratic function

A quadratic function has the following form;

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So, to get a quadratic function from the list of given options, we simply perform a comparative test of each function with the form of a quadratic function

1.\ f(x) = -8x^3 - 16x^2 - 4x

This is not a quadratic function because it follows the form f(x) = ax^3 + bx^2 + c and this is different from ax^2 +bx + c = 0 \ where \ a\neq 0

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This function has an exact match with ax^2 +bx + c = 0 \ where \ a\neq 0

By comparison; a = \frac{3}{4}\ b = 2\ and\ c = -5

3.\ f(x) = \frac{4}{x^2} - \frac{2}{x} + 1

This is not a quadratic function because it follows the form f(x) = \frac{a}{x^2} + \frac{b}{x} + c and this is different from ax^2 +bx + c = 0 \ where \ a\neq 0

4.\ f(x) = 0x^2 - 9x + 7

This is not a quadratic function because it follows the form f(x) = ax^2 +bx + c = 0\ but\ a = 0

Unlike the quadratic function where a\neq 0

So, from the list of given options, only 2.\ f(x) = \frac{3}{4}x^2 + 2x - 5 satisfies the given condition

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Tony is 4 years younger than his brother josh and two years older than his sister Cindy. Tony also has a twin brother, Evan. All
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Let Tony's age = x

He is 4 years younger than his brother Josh, so Josh's age would be x + 4

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They all add together to equal 66, so you get:

x  + x + x+4 + x-2 = 66

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4x = 64

Divide both sides by 4:

x = 64/4 = 16

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lorasvet [3.4K]

ANSWER: B. 13/8

STEP BY STEP THING:

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The function f(x)=tan(3/2x-pi/2) has:
zlopas [31]
I think the best option here is B. <span>period 2pi/3 and asymptote at x=0
</span>because period of tanx is pi so period of tan(3/2x) is pi/(3/2) = 2pi/3. I hope this can work good for you. 
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