Shifting, Reflecting, and Stretching Graphs

In order to to help with shifting, reflecting, and stretching graphs, it is important to know and understand the graphs below.

f(x)=x

f(x)=sqrt{x}

f(x)=x^2

f(x)=|x|

Graphs can be shifted horizontally and vertically using these rules:

h(x)=f(x)+k means a shift up k units

h(x)=f(x)-k means a shift down k units

h(x)=f(x+k) means a shift left k units

h(x)=f(x-k) means a shift right k units

when k is greater than or equal to 0. 

Graphs can be reflected using these rules

h(x)= -f(x) means reflection across the x-axis

h(x)= f(-x) means a reflection across the y axis

Graphs can be stretched and condensed using these rules

h(x)= c*f(x) when c>1 means  it is condensed

h(x)= c*f(x) when c<1 means it is stretched

"As a function of"

How to write functions:

A function is the relation between a set of inputs and a set of possible outputs with the property that each input has only one output.  This means that a function's input and output values are related.

When writing a function such as [f(x)] is the set of possible outputs for every input value which is represented [x]

Taking both of those statements into account, one would be able to find the relationship between many different things and define those relationships as functions.

For example: you would be able to find the relationship between this equilateral triangle's area and "x".  

The first step in doing this is to start out with the equation of what you are finding the relationship of, in this case, it is area, so you would start out with:

Where "b" is the "base" and "h" is the height. 

In this situation we know half of the overall "base"  which is "2x" so the entire base must be "4x". We know have part of our equation:

 Since this is an equilateral triangle we also know that every side is congruent or equal so that means the all sides are "4x".

This bring us to our second step. What we still don't know, is "h" or the "height" and when you are looking at the main goal, which is defining the triangle's Area as a function of "x" you can't have more than one variable (x) in there.  (That means that "h" is not allowed).

To find "h" in terms of "x" we can do one of two things.

1. We could possibly recognize this is a 30,60,90 triangle, making "h"  
2. If we couldn't figure out this is a 30,60,90 triangle, we could do the pythagorean theorem

This means it's time for step 3! We can finish our equation with only one variable so we plug that into our original equation:

Step 4 is to simplify, our first equation can be simplified to:

But wait there's more!

The last, and final step is to convert the equation into a function so we do the following:

Which tells us that "Area" or "A" as a function of "x" is equal to ""

One can write much more than just the Area as a function of "X".  One can write the perimeter of a rectangle as a function of it's area, or the volume of a package as a function of it's base, the possibilities are endless!

Review of steps:
1. Write out equation of what you are trying to find a function of
2. Make sure there is only one variable, and if there is more than one, solve in the terms of what you are looking for.  (if you are looking for A(x) then solve in terms of "x")
3. Plug in the new values for everything into your original equation
4. Simplify!
5. Convert your equation to a function.

Difference Quotients

A Difference Quotient is the formula used to find the slope of the secant line of two points on a graph, or the tangent line of a single point on a graph.

Simply, for any function f(x) the difference quotient for that function is

With h being any nonzero value in the domain of the original function.

How does one form this formula?

First take a simple, curvy graph and pick any point, lets call this point (a, f(a)) with "a" being the x value of the point, and "f(a)" being the output, or y value.
Then add any value "h" to "a" and plot a point on the graph there, creating the point (a+h, f(a+h)). Your graph will look something like this.

Since you now have 2 points, and therefore a line secant to this graph, let us find this lines slope by using the formula of the slope of a line.

And there it is, proven right above, in simplest terms the difference quotient is merely the formula of the slope of a line. But you can do so much more with the difference quotient than a plain slope formula, you can manipulate it, as long as you have a knowledge of limits, to find the slope of the tangent line of any point on a graph. Although that is for next year in calculus, for now you only need the basic knowledge of how it is formed and how it could possibly be applied.

Also here are some good tips.

• If you get 1, you probably did something very wrong.
• Stuff is going to cancel each other out in the numerator, so if it doesn't happen you distributed wrong in the f(x+h)
• It's going to divide so that there is no denominator, if it doesn't you distributed wrong again
• And as always, if it just looks funky double check your work!

By Jacob D on Nick C's account because mine was being weird.

General Functions

General Functions

Key Terms

Function: A relation in which every input has one output.

Domain: The set of inputs in a function.

Range: The set of outputs in a function.

Representations of Functions and Non-Functions

Functions

Every input (# of days) has only
one output (amount of money).

Each input for the domain has
only one output.
Ex: (-3,3), (-2,6), (0,15)

Non-Functions

When the value 5 is an input, there
are three outputs, (4, 5, 7). In addition, some input values,
3 and 4, do not have any corresponding output
values.  Therefore, this is not a function.    

The Vertical Line Test

• Used to determine if a relation is a function.
• The graph is a function if no vertical lines can be drawn that intersect the graph at more than one point.

Chapter P

Sections P.4 and P.5 were review of topics, mostly from Algebra 2, with which you should be familiar.  Section P.4 focused on solving equations of many types.  The ones we spent the most time with were equations involving fractions and radical expressions.

Perhaps the easiest way to solve an equation with rational expressions (i.e., fractions) is to multiply both sides of the equation by the least common multiple of all the denominators.   Multiplying through will eliminate all the denominators, so there will be no fractions remaining.  From there, the equation will usually be either quadratic or linear and should be relatively straightforward to solve.  Remember to check that none of your solutions make any denominator in the original equation equal zero.

When a variable is under a radical, both sides of the equation will need to be squared.  Before doing this, isolate the radical term.  (When there are two radical terms, it is typically easier to separate them before squaring both sides for the first time.)  After squaring both sides of the equation, you will usually be left with a quadratic or linear equation to solve.  Remember to check for extraneous solutions!  When you square both sides of an equation, you open the door to extraneous solutions, so you have to check them by plugging them into the original equation.

P.5 dealt primarily with absolute value and inequalities.

Definition of absolute value:

     |x| = a     if and only if     x = a  or  - x = a

Another way of defining the absolute value is to use a piecewise-defined function:
 




For inequalities:
     |x| < a     if and only if     x < a  or  - x < a

The algebraic implication of the definition of absolute value is that you should focus on what the absolute value bars are doing to the expression inside them.  And what do the absolute value bars do?  Well, it depends on the value of what's inside them.  If the quantity inside the bars is positive, then the absolute value bars leave it unchanged (i.e., the absolute value bars do nothing to what's inside them).  If the quantity inside the bars is negative, then the absolute value bars take the opposite of that quantity (i.e., it multiplies the quantity by negative one).

Example:

Once the absolute value expression is isolated on one side of the inequality sign, split the problem into two separate inequalities (based on the definition above).  If 1 - 2x is positive, then we can write the inequality without the absolute value bars.  If 1 - 2x is negative, then the absolute value bars will have the effect of multiplying that expression by negative one.  In both cases, the less than or equal to five remains unchanged.

Using interval notation, we would represent our solution as [2,3].

Polynomial inequalities are a big part of P.5. Solving them is more complicated than solving equations.
Before the official solving begins, you must have a zero on one side of the inequality.
The first to solve a polynomial inequality is to find the zeros of the polynomial.
Use these zero to set up intervals, and then pick a test value in each interval.
Plug each test value into the polynomial and see if its value is positive or negative.
Your solution will include all the intervals that matched your inequality (+ for > 0 and - for < 0).
This process makes more sense when you consider the graph of your inequality.  The inequality is > 0 when its graph is above the x-axis and < 0 when its graph is below the x-axis.  Setting up the test intervals is a numeric way of making that determination.

Example:
The first order of business is to find the zeros of the polynomial.

Next, plot the zeros on a number line to establish the test intervals, then choose a test value from each interval.  Plug that value into the polynomial and determine if the polynomial's value is positive or negative.

We were interested in where the polynomial was less than zero, so we choose the interval where f (x) was negative. 

Using interval notation, we would express our solution as (3,5).

Below is the graph of the polynomial.  Looking at that, it is clear that the interval where the graph is below the x-axis (y < 0) is from 3 to 5.