A Gallery of Graphs...or... A Family of Functions

The Rectangular Hyperbola: y=(x^(-1)), or y=(1/x)

Part 1: Introduction

A rectangular hyperbola is defined by the general rule y =(a/(x-h))+k , this is called polynomial or standard form.  The simplest hyperbola is y=(1/x) and looks like the graph below. Use Shift-Enter to execute commands.

Plot [(1/x), {x, -10, 10},PlotRange->{-5,10}]

What does each part of the ‘plot’ command above do?

Part 2: Investigate the features of rectangular hyperbolas in the form y=(a/x)

Use the red space below to create the following graphs. The first one is done for you.  
y=(1/x), y=(2/x), y=(1/(2x)), y=((-1)/x), y=((-3)/x)
Copy the command and replace (1/x)with whatever graph you are creating.

Plot [(1/x),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

What happens when you change the value of ‘a’? Complete the sentences below:
* When a is negative, 
* When a is < 1, 
* When a is >1,

Part 3: Investigate the features of rectangular hyperbolas in the form y=(a/(x-h))+k

Use the red space below to create the following graphs. The first one is done for you.  
y=(1/x), y=(1/(x-2)), y=(1/(x+2)), y=((-1)/(x-2)), y=((-1)/(x+2))
Copy the command and replace (1/x)with whatever graph you are creating.

Plot [(1/x),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

What happens when you change the value of ‘h’? Complete the sentences below:
* When h is negative, 
* When h is positive, 
* An asymptote exists at x= ?

Use the red space below to create the following graphs. The first one is done for you.  
y=(1/x), y=(1/x)-2, y=(1/x)+2, y=((-1)/x)-2, y=((-1)/x)+2
Copy the command and replace (1/x)with whatever graph you are creating.

Plot [(1/x),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

What happens when you change the value of ‘k’? Complete the sentences below:
* When k is negative, 
* When k is positive, 
* An asymptote exists at y= ?

The Truncus: y=(x^(-2)), or y=(1/(x^2))

Part 1: Investigate the features of truncus in the form y=(a/(((x-h))^2))+k

Use the red space below to create the following graphs. The first one is done for you.  
y=(1/(x^2)), y=(2/(x^2)), y=(1/(2(x^2))), y=((-1)/(x^2)), y=((-3)/(x^2))
Copy the command and replace (1/x)with whatever graph you are creating.

Plot [(1/(x^2)),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

What happens when you change the value of ‘a’? Complete the sentences below:
* When a is negative, 
* When a is < 1, 
* When a is >1,

Use the red space below to create the following graphs. The first one is done for you.  
y=(1/(x^2)),y=(1/(((x-3))^2)), y=(1/(((x+3))^2)), y=((-1)/(((x+4))^2)), y=((-1)/(((x+2))^2))
Copy the command and replace (1/x)with whatever graph you are creating.

Plot [(1/(x^2)),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

What happens when you change the value of ‘h’? Complete the sentences below:
* When h is negative, 
* When h is positive, 
* An asymptote exists at x= ?

Use the red space below to create the following graphs. The first one is done for you.  
y=(1/(x^2)), y=(1/(x^2))-2, y=(1/(x^2))+2, y=((-1)/(x^2))-2, y=((-1)/(x^2))+2
Copy the command and replace (1/x)with whatever graph you are creating.

Plot [(1/(x^2)),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

What happens when you change the value of ‘k’? Complete the sentences below:
* When k is negative, 
* When k is positive, 
* An asymptote exists at y= ?

The graph of y=(Sqrt[x]), or y=(x^(1/2))

Part 1: Investigate the features of y=(Sqrt[x]) in the form of y=a(Sqrt[x-h])+k

Plot the following graphs:
y= (Sqrt[x]), y=(Sqrt[x-3]), y=(Sqrt[x+4]), y=(Sqrt[x])-3, y=(Sqrt[x])+4, y=(Sqrt[x-3])+4, y=(Sqrt[-x])

Plot [(Sqrt[x]),{x, -10, 10},PlotRange->{-5,10}]

Plot[{"function1","function2","function3"},{x,"Minimum","Maximum"},PlotLegends->"Expressions"]

In which graphs did you alter the ‘h’ term?
In which graphs did you alter the ‘k’ term?
What is special about the point (h,k)?

Circles (not a function!)

Part 1: Investigate the features of circles in the form (((x-h))^2)+(((y-k))^2)=(r^2)

To plot circles, we need to use the command below!
Looks tricky?
It isn’t. 
Locate the centre by observing  your equation (if it is in the form above). The centre is given by the coordinates (h,k)
Identify the radius. Make sure you take the square root of the number after the equals sign to give you the radius. 
E.g. To plot (((x+1))^2)+(((y-2))^2)=4, 
 we notice that the CENTRE is given by (-1,2), and the RADIUS is (Sqrt[4]) ==2. Note that you have to put brackets around any negative coordinates.

Graph the following: (((x-1))^2)+(((y-2))^2)=9, (x^2)+(((y-3))^2)=4

ContourPlot[{"fisrt circle","second circle","third circle"},{x,"Minimum","Maximum"},{y,"Minimum","Maximum"},PlotLegends->"Expressions"]

Summarise your findings about circles:

Summarise what is similar and what is different in the different graphs studied in this unit.

Mr Barry’s Ultimate Manipulate Command!

The Manipulative below is set up to display six plots (with fixed viewing windows) so that we can see what happens as various transformations are simultaneously applied to a set of basic functions. The basic functions are:

f(x) = x		f(x) = (x^2)		f(x) = (x^3)
f(x) = (Sqrt[x])		f(x) = (1/x)		f(x) = (1/(x^2))

The transformed function is     a*f(n(x - h)) + k

Move the sliders for a, n, h and k (individually) and see how the plots change. You should notice the same effect on each graph as you change the position of the slider. 
If you want to see the actual value for a slider, click on the grey plus symbol near the slider, which opens the "animation controls".

Manipulate[Grid[{
{Plot[a(n(x-h))+k,{x,-5,5},PlotRange->5],Plot[a(((n(x-h)))^2)+k,{x,-5,5},PlotRange->5],Plot[a(n((((x-h)))^3))+k,{x,-5,5},PlotRange->5]},{Plot[a(Sqrt[n(x-h)])+k,{x,-5,5},PlotRange->5],Plot[a((1/(n(x-h))))+k,{x,-5,5},PlotRange->5],Plot[a((1/(((n(x-h)))^2)))+k,{x,-5,5},PlotRange->5]}}],
{{a,1},-5,5},{{n,1},-5,5},{{h,0},-5,5},{{k,0},-5,5}]