Question

what are the amplitude phase pellet and period of f(x)=-7sin(4x-pi)+2

Answer 1

\(\color{#000000}{ \displaystyle f(t)=A\sin(\omega t+\phi) }\)

\(\color{#000000}{ \displaystyle A }\) - amplitude
\(\color{#000000}{ \displaystyle 2\pi/\omega }\) - period
\(\color{#000000}{ \displaystyle \phi }\) - phase shift

Answer 2

And in general, whenever you have a parent function,
\(\color{#000000}{ \displaystyle y=f(t) }\), and you then add a constant \(c\) to that. i.e.,
\(\color{#000000}{ \displaystyle y=f(t)+c }\), then the new function is shifted by \(c\)
units vertically (if \(c>0\) then up, if \(c<0\) then down).

Answer 3

im lost

Answer 4

im not good at this at all can you show how its done

Answer 5

I can give you an example if you like. Ok?

Answer 6

ok

Answer 7

Actually before posting an example, I want to correct myself.

The amplitude is |A| (the absolute value of A, so if A is -5, then amplitude is just 5)

Answer 8

whats the formula for it

Answer 9

I gave you the formula for it

Answer 10

a=7?

Answer 11

oh sorry lol

Answer 12

yes, very good!
That would be the amplitude in your case

Answer 13

Can you try the period and the phase shift as well?
(I will correct you if anything)

Answer 14

ok hold on

Answer 15

Sure, take your time:)

Answer 16

is 2pi over 4x the period

Answer 17

4x?

Answer 18

\(\color{#000000}{ \displaystyle f(x)=A\sin(\omega x+\phi) }\) has a period \(2\pi/\omega\).

Answer 19

could you explain to me sorry

Answer 20

so, \(\color{#000000}{ \displaystyle f(x)=-7\sin(4 x-\pi)+2 }\) has a period of \(2\pi/4\)

Answer 21

see how these two are similar?

Answer 22

yes so thats the period

Answer 23

Yes, and you can simplify that in this case.
\(2\pi/4=\pi/2\), since \(2\) and \(4\) can be divisible by 2.

Answer 24

Ok, good with amplitude and period so far?

Answer 25

yes

Answer 26

i need the phase shift

Answer 27

Ok.

Answer 28

I made a too quick judgement, I have another error in my formula. THis error relates to my form of phase shift.

Answer 29

oh ok lol

Answer 30

In general, if you have a parent function.
\(\color{#000000}{ \displaystyle y=f(x) }\)

then to shift it c units right (or \(c\) units left if \(c\) is negative),
\(\color{#000000}{ \displaystyle y=f(x-c) }\)

Answer 31

c is -pi

Answer 32

That means that if you have a function in a form
\(\color{#000000}{ \displaystyle y=A\sin(bx) }\)

has a phase shift of \(c\) units right, of the following form:
\(\color{#000000}{ \displaystyle y=A\sin(b(x-c)) }\)

(Not just \(\color{#000000}{ \displaystyle y=A\sin(bx-c) }\) as I previously claimed)

Answer 33

So, in your case, you have:
\(\color{#000000}{ \displaystyle y=-7\sin(\color{red}{4x-\pi} ) +2 }\)

you will have to re-write the part in red, in a \(b(x-c)\) form
(because as it is now, it is in a form of \(bx-c\).)
(and that's is very different, right?)

Answer 34

so pi +4x

Answer 35

no4(x-pi)

Answer 36

\(\color{#000000}{ \displaystyle 4x-\pi=4\cdot x -4\cdot \frac{\pi}{4}=4\left(x-\frac{\pi}{4}\right) }\)

Answer 37

so whats the phase shift then? thats confusing lol

Answer 38

So, by factoring I can obtain that
\(\color{#000000}{ \displaystyle f(x)=-7\sin(4x-\pi ) +2 }\)

is the same thing as
\(\color{#000000}{ \displaystyle f(x)=-7\sin\left[4\left(x-\frac{\pi}{4}\right)\right] +2 }\)

Answer 39

4(x-pi/4)?

Answer 40

remember that y=f(x-c) is phase shifted by c units from y=f(x).

Answer 41

So, this "c" in your case is ?

Answer 42

pi

Answer 43

try again.

\(\color{#000000}{ \displaystyle f(x)=-7\sin\left[4\left(x-\frac{\pi}{4}\right)\right] +2 }\)

Answer 44

What do we subtract from \(x\) ?

Answer 45

-pi/4

Answer 46

so thats the shift?

Answer 47

Yes, be we subtract just \(\pi/4\) .

Answer 48

So the phase shift is \(\pi/4\)

Answer 49

ahhh thank you so much could you help on another one ?

Answer 50

isnt the pi negat

Answer 51

negative

Answer 52

if sin theta =3/4 what is the vale of cos theta?

Answer 53

It might be confusing often, but this is what you need to know.

A function in a form
\(\color{#000000}{ \displaystyle f(x)=A\sin(\omega[t-\phi]) }\)

has an amplitude |A|,
phase shift \(\phi\),
and a period \(2\pi/\omega\)

Answer 54

ohhh ok

Answer 55

No it is not negative.

Answer 56

thank you could you help on the one above please

Answer 57

\(\sin \theta = \frac{3}{4}\)

Answer 58

sin = opp / hyp

correct?

Answer 59

yes

Answer 60

Yes, so using the pythagorean theorem (\(a^2+b^2=c^2\)),
can you find the adjacent side?

(Hint: you know that \(\rm Opposite~Side=3\), and \(\rm Hypotenuse=4\))

Answer 61

so c^2=3^2+4^2

Answer 62

Be careful, although nice try.
"c" is always the hypotenuse!

So, you should rather say,
\(3^2+b^2=4^2\)

Answer 63

oh ok

Answer 64

Ok, can you find \(b\) ?

Answer 65

9+b^2=16 subtract 9

Answer 66

b^=7

Answer 67

you mean \(b^2=7\).
Right?

Answer 68

yea

Answer 69

And if \(b^2=7\), then \(b=?\)

Answer 70

(Hint: Take the square root of both sides)

Answer 71

b=

Answer 72

If you want to type a square root of a number c, you can use:
`\(\sqrt{~ c ~} \)` (~ is ajust a space in latex, and you can copy the code).

Answer 73

2.64

Answer 74

2.6\(\color{red}{45}\)7 --> rounds to 2.6\(\color{blue}{5}\)

Answer 75

ok thank you

Answer 76

So, if you want to approximate the square root, instead of leaving it as an exact number, you can certainly put \(2.65\).

Answer 77

We have only found the adjacent side.

Answer 78

oh

Answer 79

Now you know that the adjacent side is 2.65
and you were given that the hypotenuse is 4.

So, knowing that \(\cos \theta = \displaystyle \frac{\rm adjacent~side}{\rm ~hypotenuse} \)

Answer 80

2.65/4?

Answer 81

you will therefore get that \(\cos \theta = ?~\)

yes, and that is correct

Answer 82

0.6625

Answer 83

you can re-write it otherwise.... yes, that would also suffice :)

Answer 84

thank you so much

Answer 85

Not a problem:)

Answer 86

@areeb50: Next time, please leave the word "pellet" out of your problem statement. What did it gain you to include that the first time?

Answer 87

@mathmale it was an accident ._. english isnt my first language