21.9k views
18 votes
100 points!!! Pre calculus. I need helpppppppppp

100 points!!! Pre calculus. I need helpppppppppp-example-1
User Deewilcox
by
8.4k points

2 Answers

5 votes

Answer:

We have function,


y = 3 - 6 \sin {}^{} (2x + (\pi)/(2) )

Standard Form of Sinusoid is


y = - 6 \sin(2x + (\pi)/(2) ) + 3

Which corresponds to


y = a \sin(b(x + c)) + d

where a is the amplitude

2pi/b is the period

c is phase shift

d is vertical shift or midline.

In the equation equation, we must factor out 2 so we get


y = - 6(2(x + (\pi)/(4) )) + 3

Also remeber a and b is always positive

So now let answer the questions.

a. The period is


(2\pi)/( |b| )


(2\pi)/( |2| ) = \pi

So the period is pi radians.

b. Amplitude is


| - 6| = 6

Amplitude is 6.

c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.

So using that rule, our range is [6+3, -6+3]= [9,-3] So our range is [-3,9].

D. Plug in 0 for x.


3 - 6 \sin((2(0) + (\pi)/(2) )


3 - 6 \sin( (\pi)/(2) )


3 - 6(1)


3 - 6


= - 3

So the y intercept is (0,-3)

E. To find phase shift, set x-c=0 to solve for phase shift.


x + (\pi)/(4) = 0


x = - (\pi)/(4)

Negative means to the left, so the phase shift is pi/4 units to the left.

f. Period is PI, so use interval [0,2pi].

Look at the graph above,

100 points!!! Pre calculus. I need helpppppppppp-example-1
User DaveA
by
8.0k points
0 votes

Answer:

a. π

b. 3

c. Domain = (-∞, ∞)

Range = [-3, 9]

d. (0, 3)

e. π/4

f. See attachment 1.

Explanation:

Standard form of a sine function


\text{f}(x)=\text{A} \sin \left(\text{B}\left(x+\text{C}\right)\right)+\text{D}

where:

  • |A| = Amplitude (height from the mid-line to the peak).
  • 2π/B = Period (horizontal distance between consecutive peaks).
  • C = Phase shift (horizontal shift - positive is to the left).
  • D = Vertical shift (y=D is the mid-line of the function).

Given function:


y=3-6 \sin \left(2x+(\pi)/(2)\right)

Rearrange to standard form:


y=-6 \sin \left(2\left(x+(\pi)/(4)\right)\right)+3

Therefore:

  • |A| = 6
  • B = 2
  • C = π/4
  • D = 3

Part (a)


\sf Period & = (2 \pi)/(B)= (2 \pi)/(2)=\pi

Part (b)

Amplitude = 6

Part (c)

The domain of a function is the set of all possible x-values.

As a sine function is continuous, the domain is unrestricted: (-∞, ∞).

The range of a function is the set of all possible y-values.

As the vertical shift of the function is 3, the mid-line of the function is y=3.

As the amplitude is 6, the range is restricted to 6 more or 6 less than the mid-line: [-3, 9].

Part (d)

The y-intercept of the graph is when x = 0.

Substitute x = 0 into the function and solve for y:


\implies y=3-6 \sin \left(2(0)+(\pi)/(2)\right)


\implies y=3-6 \sin \left((\pi)/(2)\right)


\implies y=3-6 \sin \left(1\right)


\implies y=3-6


\implies y=-3

Part (e)


\sf Phase \; shift=(\pi)/(4)\;(to\;the\;left)

Part (f)

The x-intercepts of the graph are when y = 0.

Substitute y = 0 into the function and solve for x:


\implies 3-6 \sin \left(2x+(\pi)/(2)\right)=0


\implies -6 \sin \left(2x+(\pi)/(2)\right)=-3


\implies \sin \left(2x+(\pi)/(2)\right)=(1)/(2)


\implies 2x+(\pi)/(2)=\sin^(-1) \left((1)/(2)\right)


\implies 2x+(\pi)/(2)=(\pi)/(6)+2\pi n, \;\;(5\pi)/(6)+2\pi n


\implies 2x=-(\pi)/(3)+2\pi n, \;\;(\pi)/(3)+2\pi n


\implies x=-(\pi)/(6)+\pi n, \;\;(\pi)/(6)+\pi n

See attachment 1 for the sketch of the graph.

100 points!!! Pre calculus. I need helpppppppppp-example-1
100 points!!! Pre calculus. I need helpppppppppp-example-2
User Jeff Nyman
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories