Question

Graphing Calculator Problem (I don't have one): http://prntscr.com/4cv25q

Answer 1

@jdoe0001

Answer 2

are you supposed to graph it?

Answer 3

You don't have to graph it, but I'm not too sure about the entire problem.

Answer 4

hmm have you covered functions transformations yet? there seems to be an assumption you have by now

Answer 5

I have covered it, but not in detail. I don't remember the amplitude/period and where the numbers are in the equation.

Answer 6

http://maths.nayland.school.nz/Year_12/AS_2.4_Trigonometry/Images_E/Poster_images/ScreenShot059.gif <--- does that help?

Answer 7

First, @jdoe0001 , which # is the period? If I have the period, I can find the Xmin and Xmax without a problem.

Answer 8

right.... notice the picture... the "cycle" would be the period of the transformed function

Answer 9

\(\large {y=5+3cos\left(x-\frac{\pi}{3}\right)
\\ \quad \\

\begin{array}{cccccllllll}
y=&3cos(&{\color{blue}{ 1}}x&-\frac{\pi}{3})&+5\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&amplitude&period&h-shift&v-shift\\
&\downarrow &\downarrow &\downarrow &\downarrow \\
3&\frac{2\pi}{{\color{blue}{ 1}}}&\frac{\pi}{3}\ to\ right&+5\ up
\end{array}
}\)

Answer 10

hmmm lemme fix that

Answer 11

\(\large {
y=5+3cos\left(x-\frac{\pi}{3}\right)
\\ \quad \\

\begin{array}{cccccllllll}
y=&3cos(&{\color{blue}{ 1}}x&-\frac{\pi}{3})&+5\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&amplitude&period&h-shift&v-shift\\
&\downarrow &\downarrow &\downarrow &\downarrow \\
&3&\frac{2\pi}{{\color{blue}{ 1}}}&\frac{\pi}{3}\ to\ right&+5\ up
\end{array}
}\)

but yeap, notice the period

Answer 12

hmmm I'm missing a 2 there...

Answer 13

This is what I put for the answer @jdoe0001 :
Xmin = 0
Xmax = 4π

The period of this function is 2π/1 or 2π.
Since the horizontal stretching is determined by the period, and the normal period for a basic trig function is 2π, you need at least 4π (two times 2π) to contain at least two periods of the function.

Answer 14

well... let us add the 2

Answer 15

\(\large {
y=5+3cos2\left(x-\frac{\pi}{3}\right)
\\ \quad \\

\begin{array}{cccccllllll}
y=&3cos(&{\color{blue}{ 2}}x&-\frac{2\pi}{3})&+5\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&amplitude&period&h-shift&v-shift\\
&\downarrow &\downarrow &\downarrow &\downarrow \\
&3&\frac{\cancel{ 2 }\pi}{{\color{blue}{ \cancel{ 2 }}}}&\frac{2\pi}{3}\ to\ right&+5\ up
\end{array}
}\)

kinda missed the 2 before

Answer 16

Oh, so the period is π, and the answer for Xmax should be 2π

Answer 17

so that's the period for that cosine transformed function.....
so two periods will be that twice :)

recall that cosine starts off at 0 degrees

Answer 18

yeap

Answer 19

Better?
http://prntscr.com/4cvdxm

Answer 20

Thanks for the help, dude/dudet! You saved my day... and my grade!

Answer 21

hmmmm well... ... notice..... the vertical shift is not required for you in this case
since you're only asked for the x-coordinates

however the horizontal shift is

notice the cos(x) is shifted horizontally TO THE RIGHT.... so to find the Xmin and Xmax
you'd need to shift the original cos(x) THAT MUCH

Answer 22

so one can say that the \(\bf Xmin\to 0+\cfrac{2\pi}{3}\qquad Xmax\to 2\pi+\cfrac{2\pi}{3}\)

Answer 23

that'd be the x-coordinates once you push them over to the right by that much