how to graph these functions

[glass garden]

how to graph these functions

[lone prawn]

i had the same problem but i couldnt manage to do it

[stone hollow]

☹️

[sand sedge]

do you still need help?

[stone hollow]

yeah

[sand sedge]

Ok bro we can start

First we can take a look at 2cos(x)|sin(x)|

We can get rid of the absolute value

[stone hollow]

yeah it will be f(x) = 2cos x sinx and f(x) = -2cosxsinx

[sand sedge]

So ultimately the question is for which x is sin(x) ≥ 0 and for which is sin(x) < 0

[sand sedge]

yes

now the important thing is for which domain?

For which x is sin(x) ≥ 0

Look at the sinus graph

When is it positive or 0 ?

[stone hollow]

x > 0 and x< π

[sand sedge]

yes

actually also for x = 0 and x = π

[stone hollow]

and sin(x) < 0 for x> π and x< 2π

[sand sedge]

So we get f(x) = 2cos(x)sin(x) for 0 ≤ x ≤ π

[sand sedge]

yes perfect

We get:

f(x) = 2cos(x)sin(x) ; 0 ≤ x ≤ π
f(x) = -2cos(x)sin(x) ; π < x < 2π

Now we can use an identity

sin(2x) = 2cos(x)sin(x)

So we get:

f(x) = sin(2x) ; 0 ≤ x ≤ π
f(x) = -sin(2x) ; π < x < 2π

In this should be easy to draw, we would just have to figure out the sinus' period

any ideas?

still there?

[stone hollow]

yeah

im thinking

[sand sedge]

Ok another question what default period does sin(x) has?

[stone hollow]

2π?

[sand sedge]

yes

correct

now what do you think happens with sin(2x)

what consquence has the 2?

[stone hollow]

we can divide by 2?

[sand sedge]

noo

sin(2x) is faster than sin(x)

the 2 makes sin(2x) finsih twice faster than sin(x)

We can also calculate the period with the formula:

$$P=\frac{2\pi}{B}$$ where B is the stretch factor and p the period

Notice: y = Asin(B(x-C)) + D

[opal tundra]

adonhs

[sand sedge]

So what is the stretch factor of sin(2x)?

it's 2

it's the factor before x basically

[stone hollow]

okay

it makes sense

[sand sedge]

So plugging it in gives us a period of P = 2π/2 = π

yess

sin(x) has a period of 2π

sin(2x) a period of π

it finishes twice as fast

this is what i meant

So you would basically draw a sinus from 0 to pi and a .sinus from pi to 2pi

this is it

you can show a picture of your graph

[stone hollow]

okay wait a second

[sand sedge]

that's the default sin(x) graph..

[stone hollow]

so how to draw from 0 to pi and a sinus from pi to 2pi?

[sand sedge]

From 0 to pi you draw the complete sinus as we said.
From pi to 2pi you draw a negative complete sinus

[sand sedge]

So actually instead of 2pi you write pi then you got the first part

[stone hollow]

i need to do 2 graphs?

[sand sedge]

yes

that's why we did

f(x) = sin(2x) ; 0 ≤ x ≤ π
f(x) = -sin(2x) ; π < x < 2π

[stone hollow]

so first from 0 to pi and second from pi to 2pi?

[sand sedge]

You can always split a function with an absolute value in two pieces

[sand sedge]

yes but be careful which

[stone hollow]

i think i got it

[sand sedge]

:/

[stone hollow]

no?

[sand sedge]

from 0 to pi complete whole sinus in that interval

sin(2x) finishes at pi already

and the same goes from pi to 2pi

the** complete whole **negative sinus in that interval

you almost got it i believe in you

and you have to draw them together in one coordinate system

[stone hollow]

?

[sand sedge]

yes where is the other part on the coordinate system pi to 2pi

gotta draw them together eventually

[stone hollow]

do you mean where is negative pi?

[sand sedge]

no

the graph from pi to 2pi which is -sin(2x)

you almost got it

you need to draw from pi to 2pi the again a whole but negative sinus

a negative sinus is reflected at x-axis

[sand sedge]

nope

from pi to 2pi

it's negative sinus

so it's starts from below

[sand sedge]

yes

perfect

[stone hollow]

and its all for it?

[sand sedge]

yes

can you see it

[stone hollow]

okay finally

[sand sedge]

once you grasped and mastered the concepts it's actually not that hard

but it's normal

i think there another one

[stone hollow]

yeah it was a new subject for me

[sand sedge]

$$\ f(x)=\frac{1}{3}\bigg(3^{-x} -9\bigg)$$

[opal tundra]

adonhs

[sand sedge]

you are a good learner keep going

[stone hollow]

i think we should maybe multiply?

[sand sedge]

yes

$$=\frac{3^{-x}}{3}-3$$

[opal tundra]

adonhs

[stone hollow]

can we make it 3^-x-1 -3?

[sand sedge]

yesssss

very nice

[stone hollow]

but what after?

[sand sedge]

$$={3^{-x-1}-3$$

[opal tundra]

adonhs
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sand sedge]

I will tell

I would rewrite is as this

$$=3^{-(x+1)}-3$$

[opal tundra]

adonhs

[sand sedge]

Now we can read three properties

Let's make it backwards

$3^x$

[opal tundra]

adonhs

[sand sedge]

Do you know how that function looks like?

[stone hollow]

yeah

[sand sedge]

here to have this in mind

[sand sedge]

Ok now let's do this

$3^{x+1}$

[opal tundra]

adonhs

[sand sedge]

what happened to the graph any ideas

[stone hollow]

we move it

[sand sedge]

which side and by what units

[stone hollow]

right, 1

[sand sedge]

trap

it's actually left

$3^{x+1} = 3^{x-(-1)}$

[opal tundra]

adonhs

[stone hollow]

oh okay

[sand sedge]

Notice:

we also shit a function with $(x-a)$

So if we wanna shift it left a must be negative:

Example 1 (shift left):

a = -3

So:

$\ (x-(-3)) = (x+3)$

Example 2 (shift right):

a = 1

So:

$\ (x-(1)) = (x-1)$

[stone hollow]

okay i understand

[opal tundra]

adonhs

[sand sedge]

ok very good

Now we do this:

$3^{-(x+1)}$

[opal tundra]

adonhs

[sand sedge]

What does the minus do to the whole graph?

[stone hollow]

makes + -?

[sand sedge]

A minus in the exponent reflects the function at y-axis

[stone hollow]

yeah so it mirror the function

[sand sedge]

The function looks like this

_|

Now it's like this:

|_

hahaha do you understand or need a visual

[stone hollow]

understand

[sand sedge]

ok very good

almost done

[stone hollow]

and there isnt trap with -3

it goes down by 3

[sand sedge]

yes

$3^{-(x+1)} -3$

[opal tundra]

adonhs

[sand sedge]

shifts the whole function by 3 units down

[stone hollow]

and its all?

[sand sedge]

actually this functions has a root

[cerulean ruin]

yo you can use f'(x)

[sand sedge]

$3^{-(x+1)} -3 = 0\$

$\ \Rightarrow 3^{-(x+1)} = 3^1\$

$\ \Rightarrow -(x+1) = 1\$

$\ \Rightarrow x+1 = -1\$

$\ \Rightarrow x = -2\$

[opal tundra]

adonhs

[sand sedge]

So you can basically start with the graph of 3^x then draw 3^x+1 then 3^-(x+1) then 3^-(x+1) - 3

not draw all of them literally but keep in mind the changes

[stone hollow]

okay i can do it

[sand sedge]

I want to add something to this.

It was not entirely correct formulates.

$3^x$ and then $3^{-x}\$ causes a reflections at the y-axis because x = 0 meaning there is no shift yet.

[opal tundra]

adonhs

[sand sedge]

$3^{x+1}$ and then $3^{x-(-1)}\$ causes a reflection at x = -1 since the graph was moved one unit to the left

[opal tundra]

adonhs

[sand sedge]

So in general:

$a^{x-b}$ then $a^{-(x-b)}\$ reflects the function at the axis x = b

[opal tundra]

adonhs

[stone hollow]

ok i understand thanks for help

[sand sedge]

Notice an exponential functions is always greater than 0

this means there is an asymptote y = 0 that the functions approaches but never reaches

for reference

Now what do you thinks happens when the functions was shifted by negative 3 units down? Is there a new asymptote?

[stone hollow]

yes

at y=-3

[sand sedge]

perfect

[stone hollow]

and when we shift it to the left the asymptote we also shift

[sand sedge]

technically yes but no

the asymptote is a constant

a straight infinite line

so because it's infinite long it basically doesnt change

[stone hollow]

okay so we dont change it by we make another?

but*

[sand sedge]

the asymptote changes its location if the graph is shifted either up or down

$a^x > 0$

[opal tundra]

adonhs

[sand sedge]

Now $a^x + 1 > 1$

[opal tundra]

adonhs

[sand sedge]

Or $a^x -3 > -3$

[opal tundra]

adonhs

[sand sedge]

I'll send some solution of mine how it should look because I gotta go sooner or later

[stone hollow]

okay i already got it thank you very much for help

and sorry for not understanding some informations

[sand sedge]

no don't be sorry it's very normal

[stone hollow]

im trying my best

[sand sedge]

I know, you are a good and fast learner

[stone hollow]

thank you

[sand sedge]

But unfortunately being tutor here is difficult so we both try our best

I wish you very well

and if there are no questions anymore you can close this case with .solved

[stone hollow]

just type .solved?

[sand sedge]

yes

[stone hollow]

okay

.solved