Question

bros

Answer 1

dude could you help me with my question after his ?

Answer 2

I mean a general sine function have a period of 2\(\pi\)
Domain is the all values since they all go to infinity
Range is the a value
x intercept is when x is 0 (there are infinite x-int)


\(f(x)=a~sin~b((x-c))+d\)

Answer 3

good start but generally for x-intercepts they would want to describe exactly where the x-intercepts are

for sin(theta), sin(theta) = 0 when theta is an integer multiple of pi. we can express this as n*pi

Answer 4

sorry for the late reply. For domain and range how do I figure those out? i have the most trouble with those.

Answer 5

for domain, sin doesn't have any restrictions (you can plug in any x-value and get an output) so we consider the domain to be all real values

Answer 6

for range, this is something you kind of need to memorize/derive from the unit circle, but the minimum value for sin is -1 and the max is 1

Answer 7

that helps a lot!
I have another one... f(x) = sec 2x
I need help with the domain
but I think the range is (-∞,-1) (1, ∞) not sure if thats right though

Answer 8

close

since sec(2x) is basically 1/cos(2x) it does have the capacity to equal 1 or -1 so the brackets on that should be []

(-∞,-1] U [1, ∞)

Answer 9

anyway for the domain it takes all values **except*** when cos(2x) = 0

cos(theta) takes its zeros at multiples of (pi/2), so where does cos(2x) have it's zeroes?

Answer 10

pi/2? or pi/4?

Answer 11

pi/4

so basically the asymptotes will be (n)pi/4 for odd integers n

Answer 12

Oh okay and how would I say this domain?

Answer 13

“All real values except npi/4, where n is an odd integer”