Question

Help please! I will give a medal and fan!

Answer 1

Write the equation for the following graph:

Answer 2

i dont know

Answer 3

it is a little similar to graph of \(y=\tan x\)

Answer 4

The closest equation that I've found is y=3cot(x+3pi)

Answer 5

@freckles

Answer 6

\(y=3\cot(x+3\pi )\) does looks good

https://www.desmos.com/calculator/kzivodheqc

Answer 7

https://www.desmos.com/calculator/tolq44vlj4

\(y=\cot(x+3\pi)\) looks better

Answer 8

I see observe the following things:
period=pi
zero @ x=3pi/8
zero vertical shift

Answer 9

and also I see the point (pi/4,1)

Answer 10

\[y=A \cot(x +C) \\ 0=A \cot( \frac{3\pi}{8}+C) \\ 0=\cot(\frac{3\pi}{8}+C) \\ \frac{\pi}{2}=\frac{3\pi}{8}+C \\ \frac{\pi}{8}=C \\ 1=A \cot(\frac{\pi}{4}+\frac{\pi}{8}) \\ 1=A \cot(\frac{3 \pi}{8})\]
I can't figure out why it is not giving me the right graph :(
I will come back and try later when I have food in my belly

Answer 11

That's what I was doing last night and I couldn't figure it out.

Answer 12

3pi/8 wasn't the zero
I counted a bit wrong
should be the martkers are separated by (pi/4-0)/3=pi/12 units
so the zero should be at 4pi/12=pi/3

Answer 13

\[0=A \cot(\frac{\pi}{3}+C) \\ \frac{\pi}{2}=\frac{\pi}{3}+C \\ \frac{\pi}{2}-\frac{\pi}{3}=C \\ \frac{\pi}{6}=C \\ y=A \cot(x+\frac{\pi}{6}) \\ 1=A \cot(\frac{\pi}{4}+\frac{\pi}{6}) \\ 1=A \cot(\frac{5 \pi}{12}) \\ \frac{1}{ \cot(\frac{5\pi}{12})}=A \]
let me check this graph

Answer 14

i give up my numbers aren't working

Answer 15

lol maybe there is a vertical shift

Answer 16

because the 0 doesn't happen halfway between x=0 and x=pi

Answer 17

\[f(x)=A \cot(x+C)+D \\ \text{ the zero would have happened at } x=\frac{\pi}{2} \text{ but \it got shifted down 1 unit there } \\ f(x)=A \cot(x+D)-1\]

Answer 18

that D there was a type-o meant to keep it C

Answer 19

\[-1=A \cot(\frac{\pi}{2}+C)-1 \\ 0=A \cot(\frac{\pi}{2}+C) \\ \frac{\pi}{2}=\frac{\pi}{2}+C \\ C=0 \\ f(x)=A \cot(x)-1 \\ \text{ now \to find } A \\ (\frac{\pi}{4},1) \rightarrow 1=A \cot(\frac{\pi}{4})-1 \\ 2=A \cot(\frac{\pi}{4}) \\ 2=A \cdot 1\]
\[f(x)=2 \cot(x)-1\]

Answer 20

so looking at that graph I'm a lot more happier
I just did not think of a vertical shift at first
:(

Answer 21

you cool with that @kkbrookly
?

Answer 22

That equation looks very similar to the graph. Thank you! I'll use that one instead!