Question

MEDAL!!! Write the particular equation of this transformed sine graph. Assume that the horizontal shift is 1 unit to the right. (Hint: Try to find the best point to center the sinusoidal axis.)

Answer 1

@beccaboo333

Answer 2

I can't view it sorry, but @Jamierox4ev3r or @jigglypuff314 can help.

Answer 3

oh ok. Is there a reason you can't view?

Answer 4

It wouldn't show before, not it did. I don't understand this kind of math well. Sorry.

Answer 5

It's fine. Thanks for trying!

Answer 6

rather then having us download stuff to view it, we all much prefer screenshots :P
http://prntscr.com/2zgf07

Answer 7

oh sorry @jigglypuff314 thanks for screenshotting that

Answer 8

lol I forgot how to do sin graph translations even tho I just learned them last year DX sorry :/
perhaps @jdoe0001 can help?

Answer 9

haha ok thanks for trying!

Answer 10

do you know the graph for sin(x) ?

Answer 11

yes

Answer 12

notice the picture
can you see the "amplitude"? and it's new x-axis location ?
notice the vertical red line, it's indeed shifted horizontally by 1 unit

Answer 13

ok, of y=sinx the period is 2pi and the amplitude is 1

Answer 14

can you tell the period from the picture?

Answer 15

the period will be one "hump up" and one "hump down"

Answer 16

4?

Answer 17

well, the amplitude is 4, yes, 4 units up from its hub-axis and 4 down
the period is .... notice
one hump up, starts off at 1
one hump down, ends at 5
period is also 4, yes

Answer 18

and is vertically shifted by 2 units

Answer 19

y=asin(k(x-d))+c
a=4, k=90, d=1, c=2 so
y=4sin90(x-1)+2

Answer 20

because k=2pi/period

Answer 21

y=4sin90(x-1)+2 is the the answer right @jdoe0001

Answer 22

\(\bf y=Asin({\color{red}{ B}}x-{\color{blue}{ C}})+D
\\ \quad \\ \quad \\

amplitude=4\qquad period\implies \cfrac{2\pi}{B}=4\implies \cfrac{2\pi}{4}={\color{red}{ B}}\iff \cfrac{\pi}{2}
\\ \quad \\
\textit{phase shift}\implies \cfrac{C}{B}=1\implies \cfrac{C}{\frac{\pi}{2}}=1\implies {\color{blue}{ C}}=\cfrac{\pi}{2}
\\ \quad \\
\textit{vertical shift}=2
\\ \quad \\
y=Asin({\color{red}{ B}}x-{\color{blue}{ C}})+D\implies y=4sin\left({\color{red}{ \frac{\pi}{2}}}x+{\color{blue}{ \frac{\pi}{2}}}\right)+2
\\ \quad \\
\implies y=4sin\left(\frac{\pi}{2}(x+1)\right)\)

Answer 23

ahmm so much for my paste... anyhow,forgot the shift... \(\bf y=Asin({\color{red}{ B}}x-{\color{blue}{ C}})+D
\\ \quad \\ \quad \\

amplitude=4\qquad period\implies \cfrac{2\pi}{B}=4\implies \cfrac{2\pi}{4}={\color{red}{ B}}\iff \cfrac{\pi}{2}
\\ \quad \\
\textit{phase shift}\implies \cfrac{C}{B}=1\implies \cfrac{C}{\frac{\pi}{2}}=1\implies {\color{blue}{ C}}=\cfrac{\pi}{2}
\\ \quad \\
\textit{vertical shift}=2
\\ \quad \\
y=Asin({\color{red}{ B}}x-{\color{blue}{ C}})+D\implies y=4sin\left({\color{red}{ \frac{\pi}{2}}}x+{\color{blue}{ \frac{\pi}{2}}}\right)+2
\\ \quad \\
\implies y=4sin\left(\frac{\pi}{2}(x+1)\right)+2\)

Answer 24

hmmm lemme recheck.. onthe - or +.. is to the right... .should be minus I think

Answer 25

yeah x-1 is right

Answer 26

yeah.. it should be ... minus anyhow \(\bf y=Asin({\color{red}{ B}}x-{\color{blue}{ C}})+D
\\ \quad \\ \quad \\

amplitude=4\qquad period\implies \cfrac{2\pi}{B}=4\implies \cfrac{2\pi}{4}={\color{red}{ B}}\iff \cfrac{\pi}{2}
\\ \quad \\
\textit{phase shift}\implies \cfrac{C}{B}=1\implies \cfrac{C}{\frac{\pi}{2}}=1\implies {\color{blue}{ C}}=\cfrac{\pi}{2}
\\ \quad \\
\textit{vertical shift}=2
\\ \quad \\
y=Asin({\color{red}{ B}}x-{\color{blue}{ C}})+D\implies y=4sin\left({\color{red}{ \frac{\pi}{2}}}x-{\color{blue}{ \frac{\pi}{2}}}\right)+2
\\ \quad \\
\implies y=4sin\left(\frac{\pi}{2}(x-1)\right)+2\)

Answer 27

yeap

Answer 28

thanks a lot!

Answer 29

yw