ALGEBRAIC FUNCTIONS

A set of all X's is called the DOMAIN

A set of all Y's is called the RANGE

A FUNCTION is when the X value of each X,Y pair of coordinates is always a different value.  

A sample set of XY pairs of a function      { (5,2) (6,4) (7,8) }
A sample set of XY pairs "Not" a function { (5,2) (5,4) (5,8) }

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f(x) = y or in this case f(x) = 2x + 1 since y = 2x + 1
f(DOMAIN independent value) = (RANGE dependent value)

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In this example the range y = 2x + 1	This is a function, so... f(x) = 2x + 1
EXAMPLE 1 when x = - 1:	EXAMPLE 2 when x = + 2:
Given;  f(x) = 2x + 1  When;  x = -1 then;    f(-1) = 2(-1) + 1 so;         f(-1) = - 2 + 1 and;      f(-1) = -1 conclusion;  f(x) = -1 when x = -1	Given;  f(x) = 2x + 1 when;  x = 2 then;   f(2) = 2(2) + 1 so;        f(2) = 4 + 1 and;     f(2) = 5 conclusion;  f(x) = 5 when x = 2

The equation of a straight line can be written in three different forms. Below are three equation forms and proof they are all equal to each other.

y – y₁ = m(x - x₁)        :: Point slope equation form.
                                    Where m=slope and y – y₁ and  x - x₁ are points on the line
y = mx + b                :: Slope Intercept form.   
                                     Where m=slope, b=y interecpt when x=0
Ax + By = C               :: Standard form.

Example:

Lets take the straight line that crosses two (x, y) points of (3,2) and (8,4). The equations for this line are shown below.  All three forms of the equation represent the same straight line shown in the graph.

y - 2 = 2/5(x-3) :: Point slope form
y = 2/5x + .8     :: Slope intercept form
2x - 5y = - 4      :: Standard form

We now will provide proof that the line shown in the graph can be represented by an equation in three different forms. Since all the equation forms are equal to one another they can be converted from one form to another and still represent the line. Notice also that if we are given two sets of (x,y) points on the line we can calculate the line slope and y intercept. This is all the information we need to fully describe the line in any of the three equation formats.

·         Convert two points (3,2) and (8,4) to point slope equation form.

(y2 - y1) / (x 2 - x1)          :: Find the slope of the line between two points.
(4 - 2) / (8 - 3)                  :: Enter the values of the two points.
2/5                                  :: This is the slope or m of the line.
(y - 2) / (x - 3) = 2/5         :: Replace the y2 = 4 and x 2 = 8 points with y and x.
((y -2)/(x -3)) * (x -3) = 2/5 * (x -3) :: Multiply both sides by (x-3)
y - 2 = 2/5x + .8               :: Point slope form y - y1 = m(x - x1)

·         Convert the point slope form of the equation to the slope intercept form.

y - 2 = 2/5(x - 3)              :: Point slope form y - y1 = m(x - x1)
y - 2 +2 = 2/5(x - 3) +2     :: Add +2 to each side
y = 2/5*x - 2/5*3 + 2        :: Multiply the term (x-3) by 2/5
y = 2/5*x - 1.2 + 2           :: Now add -1.2 and +2
y = 2/5x - .8                    :: Change the decimal to a fraction
y = 2/5x - 4/5                  :: Slope intercept form y = mx + b

·         Convert the slope intercept form to the standard form.

y = 2/5x - 4/5                     :: Slope intercept form y = mx + b 
y * 5 = (2/5x)*5 - (4/5)*5      :: Multiply each side by 5
5y = 2x - 4                         :: Move 2x to the left side and change the sign.
5y - 2x = - 4                       :: Standard form Ax + By = C

·         Convert standard form to point slope form.

2x - 5y = - 4                  :: Standard form of Ax + By = C
2x + 4 = 5y                   :: Move the -4 to the left and 5y to the right.
5y = 2x + 4                   :: Reverse the order to get 5y on the left.
5y/5 = (2x + 4)/5            :: Convert the fraction 4/5 to a decimal
y = 2/5x + .8                 :: Point slope form. y - y1 = m(x - x1)

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·         Given: (x₁ , y₁) and (x₂, y₂) representing two points on a line;
use (y - y₁)/(x - x₁) = (y₂ - y₁) / (x₂ - x₁) to solve for the lines equation in standard form.

·         Given:  (x, y) and m representing one point on the line and the slope of the line;
use (y - y₁)/(x - x₁) = m and solve for y giving the slope intercept equation form.

·       Given: m and b representing the slope of the line and the y intercept;
use y = mx + b to give the slope intercept form of the equation.