Graphing question (1 Viewer)

[Joshmosh2]

Hello,
How do I sketch:
1. y = f(|x|)
2. y = |f(x)|
3. |y| = f(x)

For the function:
http://imgur.com/P2xfTr5

I have a couple things that need clarification:
*Is it better to find the equation of the function and somehow continue accordingly? or is it better to recognise the nature of each variant and visually change the graph
*I am unsure about where to start for questions 1 and 3

 

[seventhroot]

ans to questions:

*1) no
*2)

1) basically think of the y axis as a mirror then draw the same thing
2) everything gets flipped above the x axis
3) same thing as 2 but then mirror it across the x axis as well

 

[faisalabdul16]

You will never find the equation if they merely give you a graph

 

[seventhroot]

You will never find the equation if they merely give you a graph

yes this; even of you got the equation; it is pointless because they're going to be worth 1 or 2 marks anyways

 

[seventhroot]

and enjoy some free 3U notes: http://tinyurl.com/freeMX1notes

 

[integral95]

ans to questions:

*1) no
*2)

1) basically think of the y axis as a mirror then draw the same thing
2) everything gets flipped above the x axis
3) same thing as 2 but then mirror it across the x axis as well

if i was a newbie i'd be confused at that lol

 

[aDimitri]

Hello,
How do I sketch:
1. y = f(|x|)
2. y = |f(x)|
3. |y| = f(x)

For the function:
http://imgur.com/P2xfTr5

I have a couple things that need clarification:
*Is it better to find the equation of the function and somehow continue accordingly? or is it better to recognise the nature of each variant and visually change the graph
*I am unsure about where to start for questions 1 and 3

by learning how in class

 

[trapizi]

Let's make this simple .
1) Erase the left hand side of the graph then reflect the right hand side across the y axis
2) Fold the bottom up.
3) Delete everything that is under the x axis and reflect what's above the x axis.

 

[Joshmosh2]

Ok thanks! I have another question..
Sketch the curve
y^2 = x(x-1)

Here is how I approached it:
Firstly, y = +- sqrt(x(x-1))
so the domain would be
x <= 0 and x >=1

And after differentiating, there turns out to be no stationary points

x intercepts at 0,1
y = 0

Now here's the thing; I don't think I have enough information to work from here. What was suggested in the textbook was to draw a 'guide graph' which basically indicated the places where the graph was above and below the x-axis.

From what I can tell from the examples they gave, the function y^2 = f(x) has two main variants;
- a 'loop' that looks like a fish
- a double modulus with curved edges

But due to the domain they gave, it would sound logical that the graph would be split into two, like a double modulus


Moving that aside though, If there are no stationary points to work from, what do you do next?

 

[seventhroot]

Ok thanks! I have another question..
Sketch the curve
y^2 = x(x-1)

Here is how I approached it:
Firstly, y = +- sqrt(x(x-1))
so the domain would be
x <= 0 and x >=1

And after differentiating, there turns out to be no stationary points

x intercepts at 0,1
y = 0

Now here's the thing; I don't think I have enough information to work from here. What was suggested in the textbook was to draw a 'guide graph' which basically indicated the places where the graph was above and below the x-axis.

From what I can tell from the examples they gave, the function y^2 = f(x) has two main variants;
- a 'loop' that looks like a fish
- a double modulus with curved edges

But due to the domain they gave, it would sound logical that the graph would be split into two, like a double modulus


Moving that aside though, If there are no stationary points to work from, what do you do next?

this is how I would do it:

graph y = x(x-1) //standard year 9 question

no you want y^2 which is equiv to sqrt[ x(x-1) ] so sqrt the y values to your graph in part 1 and since it is +/- make sure to reflect it about the x axis

sqrt(x(x-1))

huehuehue

 

[Joshmosh2]

this is how I would do it:

graph y = x(x-1) //standard year 9 question

no you want y^2 which is equiv to sqrt[ x(x-1) ] so sqrt the y values to your graph in part 1 and since it is +/- make sure to reflect it about the x axis

Waow.. so simple

 

[seventhroot]

Waow.. so simple

no need to over complicate it

 

[trapizi]

Now here's the thing; I don't think I have enough information to work from here. What was suggested in the textbook was to draw a 'guide graph' which basically indicated the places where the graph was above and below the x-axis.

From what I can tell from the examples they gave, the function y^2 = f(x) has two main variants;
- a 'loop' that looks like a fish
- a double modulus with curved edges

But due to the domain they gave, it would sound logical that the graph would be split into two, like a double modulus


Moving that aside though, If there are no stationary points to work from, what do you do next?

You do have enough information to draw the +-sqrt[f(x)]. You need to analyse how a sqrt function is defined.
-There is no negative value of y in a sqrt function
-Roots of f(x) is roots of sqrt[f(x)]
-Horizontal asymptote of sqrt[f(x)] is sqrt(horizontal asymtote of f(x))

So your graph should look like this : http://www.wolframalpha.com/input/?i=plot+y=x(x-1)+and+y=-sqrt(x(x-1))+and+y=sqrt(x(x-1))

 

[integral95]

also remember

When you sqrt a value between 0 and 1 it makes it bigger,

While sqrting a value over 1 makes it smaller

ofc sqrting 1 is still 1

There are MC questions that has those comparisons

 

[Joshmosh2]

also remember

When you sqrt a value between 0 and 1 it makes it bigger,

While sqrting a value over 1 makes it smaller

ofc sqrting 1 is still 1

There are MC questions that has those comparisons

Ok cool, i'll keep that in mind