Question

Can some one show me where I went wrong on this problem. I checked my answer with a graphing calculator and my points are correct, but the calculator graphs it as -cos (inverted from what I came up with).... why?

Answer 1

\[y=Acos( \omega x - \phi)\]

A: amplitude
\(\omega\): angular velocity ( from which we can find period)
\(\phi: shift\)

you need to bring the given function to the above format,
and compare to find amplitude, angular velocity( and consequently, period) and shift.

Answer 2

did i go wrong leaving w negative?

Answer 3

w=-1

Answer 4

leaving w as a negative number is not wrong...
but it is inconvenient, and may cause you to make errors...

are you aware of this property of 'cos'
\(cos(-\theta) = cos(\theta)\)

try putting that to use, and getting rid of the negative sign with w

Answer 5

I did not know that... let me try again..

Answer 6

the period has to be a positive number

Answer 7

and the plot in your attachment is incorrect rather than the calculations

Answer 8

re-did the work with that rule and my "shift" is now (-pi/2,3pi/2)...

Answer 9

these point are right on the calculator, but there is still a -cos curve between that period... I don't see where that is coming from...

Answer 10

what do you mean (-pi/2, 3pi/2)

'shift' is a single number...

can you explain what you have understood about what 'shift' means?

Answer 11

the start and end points on the x axis for one whole period

Answer 12

i get the fist point from "ϕ" and the second point from ϕ+period

Answer 13

nope....

shift is how much the function you have is moved along the x-axis, compared with a zero shift function,

for example,

y=cos(x-pi/2)
has a shift of pi/2,

which means, y=cos(x-pi/2) is shifted to the 'right' along the x axis by pi/2 units compared with y= cos(x-0)

Answer 14

ok. i see that.... the second point I was adding is just where to end one period as we are not allowed calculators in class and usually just graph one period.

Answer 15

either way... starting at -pi/2 ... the graph is a -cos curve from there ... y?