Question

DFSGFSDGDSGS

Answer 1

i know it should be some version of then tan(x) fucntion

Answer 2

@sourwing @phi

Answer 3

@hartnn @amistre64

Answer 4

@aaronq @Luigi0210

Answer 5

@ganeshie8 @Nurali

Answer 6

yes, it is a tan function, tan(k x)
we need to find the value of k
if it showed tan(x)=1 when x = pi/4 you would be done. (k would be 1)
but you have
tan(k x) =1 when x is ½
in other words tan(k/2) = 1
we know k/2 must equal pi/4 (because tan(pi/4) =1 )
k/2 = pi/4
k= pi/4 * 2
k = pi/2

so you have
\[ \tan\left(\frac{\pi}{2}x\right) \]

Answer 7

do you think you could help me with a couple more?

Answer 8

for this one i had tan(-x-1)

Answer 9

@phi?

Answer 10

this one is trickier.

It is flipped so we should expect -kx
it is shifted so we should expect -k x + b
where b is the amount of shift

I would do it in steps: the curve is shifted left by 1 step. we use x+1 (this means when x=-1 in tan(x+1) we would get tan(-1+1)= tan(0), which is what we want

so we have tan(k (x+1))
we need to find k.
at x= 1, the curve is zooming down... that matches the normal tan function at -pi/2
in other words, k(x+1) at x=1 = -pi/2
or k*2= -pi/2
k= -pi/4

so we have tan(-pi/4(x+1) ) or
\[ \tan\left( -\frac{\pi}{4}x - \frac{\pi}{4}\right) \]

Answer 11

yes, looks good

Answer 12

my homework is online and it said i was wrong

Answer 13

I guess we have to be more careful. The curve is exactly -2 when x=1

if we start with
y= -sec(k x)
we need to find k
-sec( k x) = -1/cos( k x)
at (1,-2) we have -1/cos( k*1) = -2
or
cos(k) = ½

cos is ½ at 60º or pi/3
so k= pi/3
and your curve is -sec(pi/3 * x)

Answer 14

ok and do you know how to make sec(x) skinnier and more compact?

Answer 15

make "k" bigger
in sec(k x)

Answer 16

sorry i meant -sec(pi(x)) yeah i plugged it in its right. thanks so much

Answer 17

yes, -sec(pi*x)