Equation of Trig Graph

[/0]

The graph shown has rule of the form

Find the values of and .

I'm ok with the amplitude, but funny things happen when I try to solve for .... thanks.

[AppleXY]

It's kinda hard solving for A, when there isn't any info?

[cara.mel]

It's been been translated pi/3 to the right => e = -pi/3

[Collin Li]

When solving for , you will get a bunch of possible solutions. This is because of:
i) the same value occurring more than once in a cycle
ii) the cycle repeating every

[AppleXY]

When solving for , you will get a bunch of possible solutions. This is because of:
i) the same value occurring more than once in a cycle
ii) the cycle repeating every

my point exactly.

[plato]

The graph as drawn seems to have a maximum of 0.5 and a minimum of around -0.6 as the y-intercept is shown marked at -0.5. So there is something wrong with the diagram if the equation is to be in the form

The diagram suggests a vertical translation has occurred downward by 0.05 units or so but the required equation does not allow for this.

Therefore, one assumes the minimum value ought to be -0.5 and the y-intercept is around -0.433. This assumes the equation you are after is

There are certainly an infinite number of values for but the commonly appropriate one will be within . Otherwise, you might as well write the formula as a cosine function.

[cara.mel]

* caramel is idiot and only sees things in the way vce teaches us

Just go 2n*pi - pi/3 then or whatever. xD

[/0]

Ok this is what I had problems with



Then



But if you go with the second one you get the graph flipped?!

[Mao]

looking back, your method seems a bit shaky, as the variables have not been used so correctly...

to solve for epsilon:





this happens when [1] and

since the second solution () lies in the second/third quadrant:

(3rd quadrant) [2]or (2nd quadrant) [3]
which means
from [1]

from [2]

from [3]

same answer

you need to take in consideration of quadrants and the entire cycle, rather than just half the period of

[/0]

ok thx