Summary for Lesson 10 Graphing Reciprocal Functions

Didn't understand this lesson to the fullest hoping my classmates will explain it better.

An example of a graph of a reciprocal function looks like this http://wpcontent.answcdn.com/wikipedia/commons/thumb/2/29/Rectangular_hyperbola.svg/330px-Rectangular_hyperbola.svg.png

To find the reciprocal of a graph, you would take the original equation ex. y=2x+2 and inverse the "x" side of the equation so it becomes y=1/2x+2

After this set up a table of values for the y and x and the 1/y so that you can plot the original graph points and than its reciprocal function points.

After setting up table of values, solve for variant points by setting value of y to 1 and -1 and solving for x on both of your new equations.

Ex.) y=2x+1

1=2x+1 --> 0=2x --> x=0 y=1 (invariant point)

-1=2x+1 --> -2=2x --> x= -1, y= -1 (invariant point)

The invariant point is to see where the line of the graph will start to turn so it's acting as a turning point for the line.

After plotting invariant points and setting up table of values all thats left to do is plot the reciprocal graph

Make sure to label horizontal and vertical asymptotes and draw arrowheads for both lines, because as a function the line continues outside of the graph.

Transformations Lessons

Here are the last few lessons from our transformations unit.  Enjoy!

Transformations Lesson 2

Transformations Lesson 3

Symmetry, Reflections and Inverses - March 14, 2011

Review:

A graph is a function if all the y values have a unique x value.

If it passes the "Vertical Line Test" it means it is a function.

And a function is said to be a one-to-one function if the inverse of the graph is still a function.

Reflection of Functions:

y=f(x) is what you call your "parent graph". Points and equations will be plotted on the Cartesian plane normally.

If a negative one (-1) is added in front of the "parent" equation, so it changes from y=f(x) to y=-f(x), this means that you will have to reflect the graph over the x-axis. This means that the new points of the graph are equal to the points of the "parent" graph when you fold the plane horizontally at the x-axis.

y = f(x) >>>>>>>>>>>>>>>> y = -f(x)

x = 0 3 3 >>>>>>>>>>>>>>>> x = 0 3 -3

y = 2 3 -1 >>>>>>>>>>>>>>>> y= -2 -3 1

If a negative one (-1) is added in front of the x in your parent graph, meaning your equation changed from y=f(x) into y=f(-x), you have to reflect your graph over the y-axis. So this means that your new points would equal you parent graph's points if you were to fold the whole Cartesian plane vertically on the y-axis.

y = f(x) >>>>>>>>>>>>>>>> y = f(-x)

x = 0 3 3 >>>>>>>>>>>>>>>> x = 0 -3 3

y = 2 3 -1 >>>>>>>>>>>>>>>> y= 2 3 -1

Finally if you try to inverse functions that means your "parent" equation changed into y=f^-1(x) you just interchange your x and y values. Your new points should also equal to the parent graph's new points when you fold the Cartesian plane diagonally on y=x .

y = f(x) >>>>>>>>>>>>>>>> y = f^-1(x)

x = 0 3 3 >>>>>>>>>>>>>>>> x = 2 3 -1

y = 2 3 -1 >>>>>>>>>>>>>>>> y= 0 3 3

Finding the Inverse Equation

To find the inverse equation you need to switch the spots of the y and x in the equation.

If the equation is y=4x-7 you change it into x=4y-7 and then you have to rearrange the new equation isolating y. In this case your new equation will be y=(x+7)/3

If there is a specific value for the inverse, such as f^-1(2) , just find the inverse equation first and then substitute the 2 for the x value.

To find f^-1(2), given f(x)=2x+3

f(x)=2x+3

y=2x+3 >>>>>> y=(x+3)/2

f^-1(x)=(x+3)/2 f^-1(x) = y

f^-1(2)=(x+3)/2 >>>>>>> y=(2+3)/2

y=5/2

Therefore your f^-1(2) , in the equation f(x)=2x+3, is ( 2 , 5/2 )

That's it ! The next blogger will be ......... Nick M.

How to determine the equation of a trigonometric graph

Check out this great video for help with determining the equation of a sine or cosine function given the graph.

 

http://www.youtube.com/watch?v=yvW5l9W1hgE&feature=player_embedded - at=158

Transformations of Sine and Cosine Graphs

At the end of last week we looked at transformations of Sinusoidal graphs.  We looked at how the parameters A, B, C and D affect the position and shape of the graph.

Parameters A B C and D and what they do...

y = A sin f [B(x-C)] + D

A- Vertical Stretch/ Compression
B- Horizontal Stretch/Compression
C- Horizontal Translation
D- Vertical Translation

To find the period of the graph use P=2pi/B.

If B is attached to the x or theta value, you must factor it out of both terms inside the bracket.

For all you visual learners....Check out this great website applet to help you SEE how the parameters change the shape of the graph.  Enjoy.     http://www.ronblond.com/M12/sc.APPLET/index.html

We will continue with lessons on transformations of Cubic functions, Square Root functions and Rational functions.

See you Monday.  Monday's blogger will be.....

"Moving on" with Transformations

Today we had our first unit assessment on Circular Functions.  We will continue on tomorrow with our transformations unit.  We will be looking specifically at Trigonometric functions and their respective graphs.  We will look at how the A, B,C and D in a sinusoidal equation, affect the position and shape of the graph.  We will also look at drawing the inverses of each of these graphs.

Transformations

Today we started our second unit in Grade 12 PreCalculus on Transformations.
We looked at shifting a graph to the left or right, up or down.  When there is a horizontal shift we look for the "h" value.  The shift is opposite to the sign that is in the equation.  When there is a vertical shift we look for the "k" value.  The shift is the same as the sign that is in the equation, plus = up, minus = down.

Transformations Lesson 1 

REMINDER:  CIRCULAR FUNCTIONS UNIT ASSESSMENT  Friday March 5th, 2011 
Date changed to Tuesday March 8,2011