Question

What would a cosine function look like with a given amplitude of 7, period of pi/6, and vertical shift of down 2?

Answer 1

please just give me a straighforward answer

Answer 2

ill fan and medal:)

Answer 3

\[y=A \cos(b(x-c))+d \\ |A| \text{ is the amp number } \\ \frac{2 \pi}{b} \text{ is the period } \\ c \text{ is the phase shift number } \\ d \text { is the vertical shift } \\ \text{ you are given } |A|=7 \\ \text{ and } \frac{2\pi}{b}=\frac{\pi}{6} \\ \text{ and also what } d \text{ is }\]

Now if you want to draw this...
Assuming the phase shift is 0....
recall cos(x) is a function between -1 and 1
well you have A cos(x) which puts the function between -A and A
A cos(x)+d puts the function between -A+d and A+d
So adding A and d and the opposite of A and d give you the max and min of the function


recall the special points of y=cos(x) are
(0,1)
(pi/2,0)
(pi,-1)
(3pi/2,0)
(2pi,1)
that is since
cos(0)=1
cos(pi/2)=0
cos(pi)=-1
cos(3pi/2)=0
cos(2pi)=1

now since we have the period is a bit difference for our function that will change a bit
so say we have y=cos(bx)
how does this effect our special numbers we plug in
Well the period of y=cos(x) was 2pi
and the period for y=cos(bx) is 2pi/b

so like we had 4 spaces between our special numbers
we will have 4 spaces again
2pi/(4b)
so

cos(0)=1
cos(0+2pi/(4b))=cos(pi/(2b))=0
cos(pi/(2b)+pi/(2b))=cos(pi/b)=-1
cos(pi/b+pi/(2b))=cos(3pi/(2b))=0
cos(3pi/(2b)+pi/(2b))=cos(2pi/b)=1


so the graph will either be stretched out more or less horizontally
so
y=cos(2x) looks like

see the period is 2pi/2=pi



anyways I hopes this gives you a better idea